Mixing
How to find the value $sumlimits_{n=1}^{infty}frac{(-1)^{n}}{2n-1}$ converges to?
$begingroup$
We have been finding the value of sums related to the Harmonic series by using that
$sum_{n=1}^{N}frac{1}{n}= ln(N)+gamma + varepsilon_N$.
When I try to calculate $sum_{n=1}^{infty}frac{(-1)^{n}}{2n-1}$ using that method there is a point where I don´t know how to continue.
I have read that its value is $frac{pi}{4}$ but I don´t understand how is that obtained and I would like to know if it´s possible to solve it using the method I know.
This is what I did:
$S_N=sum_{n=1}^{N}frac{(-1)^n}{2n-1}$
$S_{2N}=sum_{n=1}^{2N}frac{(-1)^n}{2n-1}=-1+frac{1}{3}-frac{1}{5}+...-frac{1}{4N-3}+frac{1}{4N-1}=$
$=1+frac{1}{3}+frac{1}{5}+...+frac{1}{4N-1}-2(1+frac{1}{5}+frac{1}{9}+...+frac{1}{4N-3})=sum_{n=1}^{2N}frac{1}{2n-1}-2sum_{n=1}^{N}frac{1}{4n-3}$
.
$sum_{n=1}^{2N}frac{1}{2n-1}=1+frac{1}{3}+frac{1}{5}+...+frac{1}{4N-3}+frac{1}{4N-1}=$
$=1+frac{1}{2}+frac{1}{3}+...+frac{1}{4N-1}+frac{1}{4N}-(frac{1}{2}+frac{1}{4}+...+frac{1}{4N})=sum_{n=1}^{4N}frac{1}{n}-frac{1}{2}sum_{n=1}^{2N}frac{1}{n}$
.
But I don´t know what to do with the second one. I tried to write it like this:
$sum_{n=1}^{N}frac{1}{4n-3}=1+frac{1}{5}+frac{1}{9}+...+frac{1}{4N-3}=$
$=1+frac{1}{3}+frac{1}{5}+...+frac{1}{4N-3}+frac{1}{4N-1}-(frac{1}{3}+frac{1}{7}+...frac{1}{4N-1})=sum_{n=1}^{2N}frac{1}{2n-1}-sum_{n=1}^{N}frac{1}{4n-1}$
Wich didn´t help because I find $sum_{n=1}^{N}frac{1}{4n-1}$ as difficult to solve as $sum_{n=1}^{N}frac{1}{4n-3}$ and I have no idea of what else to do.
sequences-and-series summation
edited Jan 28 at 16:53
Did
249k23226466
asked Jan 28 at 16:42
ChalklingChalkling
144
$endgroup$
add a comment |
$begingroup$
We have been finding the value of sums related to the Harmonic series by using that
$sum_{n=1}^{N}frac{1}{n}= ln(N)+gamma + varepsilon_N$.
When I try to calculate $sum_{n=1}^{infty}frac{(-1)^{n}}{2n-1}$ using that method there is a point where I don´t know how to continue.
I have read that its value is $frac{pi}{4}$ but I don´t understand how is that obtained and I would like to know if it´s possible to solve it using the method I know.
This is what I did:
$S_N=sum_{n=1}^{N}frac{(-1)^n}{2n-1}$
$S_{2N}=sum_{n=1}^{2N}frac{(-1)^n}{2n-1}=-1+frac{1}{3}-frac{1}{5}+...-frac{1}{4N-3}+frac{1}{4N-1}=$
$=1+frac{1}{3}+frac{1}{5}+...+frac{1}{4N-1}-2(1+frac{1}{5}+frac{1}{9}+...+frac{1}{4N-3})=sum_{n=1}^{2N}frac{1}{2n-1}-2sum_{n=1}^{N}frac{1}{4n-3}$
.
$sum_{n=1}^{2N}frac{1}{2n-1}=1+frac{1}{3}+frac{1}{5}+...+frac{1}{4N-3}+frac{1}{4N-1}=$
$=1+frac{1}{2}+frac{1}{3}+...+frac{1}{4N-1}+frac{1}{4N}-(frac{1}{2}+frac{1}{4}+...+frac{1}{4N})=sum_{n=1}^{4N}frac{1}{n}-frac{1}{2}sum_{n=1}^{2N}frac{1}{n}$
.
But I don´t know what to do with the second one. I tried to write it like this:
$sum_{n=1}^{N}frac{1}{4n-3}=1+frac{1}{5}+frac{1}{9}+...+frac{1}{4N-3}=$
$=1+frac{1}{3}+frac{1}{5}+...+frac{1}{4N-3}+frac{1}{4N-1}-(frac{1}{3}+frac{1}{7}+...frac{1}{4N-1})=sum_{n=1}^{2N}frac{1}{2n-1}-sum_{n=1}^{N}frac{1}{4n-1}$
Wich didn´t help because I find $sum_{n=1}^{N}frac{1}{4n-1}$ as difficult to solve as $sum_{n=1}^{N}frac{1}{4n-3}$ and I have no idea of what else to do.
sequences-and-series summation
edited Jan 28 at 16:53
Did
249k23226466
asked Jan 28 at 16:42
ChalklingChalkling
144
$endgroup$
1
$begingroup$
Hint: $$sum_{n=1}^{infty}frac{(-1)^{n}}{2n-1}=sum_{n=1}^{infty}frac{a_n}{n}$$ where $(a_n)$ has period $4$ hence $$a_n=sum_{k=1}^4alpha_ki^{kn}$$ since $i$ is a principal fourth root of $1$, for some given $(alpha_1,alpha_2,alpha_3,alpha_4)$ independent of $n$. Now, it happens that $alpha_4=0$. And remains to compute $$u(i^k)=sum_{n=1}^inftyfrac{i^{kn}}n$$ for $k=1$, $k=2$ and $k=3$. Can you do that? At the end, $$sum_{n=1}^{infty}frac{(-1)^{n}}{2n-1}=alpha_1u(i)+alpha_2u(-1)+alpha_3u(-i)$$
$endgroup$
– Did
Jan 28 at 16:49
$begingroup$
We are still doing only real analysis, so sadly I can´t use that
$endgroup$
– Chalkling
Jan 28 at 17:07
$begingroup$
The list of tools you cannot use grows longer any minute, it seems (see your comment to the answer below for another example). Say, what are the tools you can use?
$endgroup$
– Did
Jan 28 at 17:12
$begingroup$
Silence... Probably not either the identity $$picot(pi z)=frac1z+2zsum_{n=1}^inftyfrac1{z^2-n^2}$$ at $z=frac14$...
$endgroup$
– Did
Jan 28 at 17:23
$begingroup$
No, not either. As I said below, right now can´t use almost anything other than the method we have been taught to use for this series and I wanted to know if that was enogh
$endgroup$
– Chalkling
Jan 28 at 17:31
add a comment |
$begingroup$
We have been finding the value of sums related to the Harmonic series by using that
$sum_{n=1}^{N}frac{1}{n}= ln(N)+gamma + varepsilon_N$.
When I try to calculate $sum_{n=1}^{infty}frac{(-1)^{n}}{2n-1}$ using that method there is a point where I don´t know how to continue.
I have read that its value is $frac{pi}{4}$ but I don´t understand how is that obtained and I would like to know if it´s possible to solve it using the method I know.
This is what I did:
$S_N=sum_{n=1}^{N}frac{(-1)^n}{2n-1}$
$S_{2N}=sum_{n=1}^{2N}frac{(-1)^n}{2n-1}=-1+frac{1}{3}-frac{1}{5}+...-frac{1}{4N-3}+frac{1}{4N-1}=$
$=1+frac{1}{3}+frac{1}{5}+...+frac{1}{4N-1}-2(1+frac{1}{5}+frac{1}{9}+...+frac{1}{4N-3})=sum_{n=1}^{2N}frac{1}{2n-1}-2sum_{n=1}^{N}frac{1}{4n-3}$
.
$sum_{n=1}^{2N}frac{1}{2n-1}=1+frac{1}{3}+frac{1}{5}+...+frac{1}{4N-3}+frac{1}{4N-1}=$
$=1+frac{1}{2}+frac{1}{3}+...+frac{1}{4N-1}+frac{1}{4N}-(frac{1}{2}+frac{1}{4}+...+frac{1}{4N})=sum_{n=1}^{4N}frac{1}{n}-frac{1}{2}sum_{n=1}^{2N}frac{1}{n}$
.
But I don´t know what to do with the second one. I tried to write it like this:
$sum_{n=1}^{N}frac{1}{4n-3}=1+frac{1}{5}+frac{1}{9}+...+frac{1}{4N-3}=$
$=1+frac{1}{3}+frac{1}{5}+...+frac{1}{4N-3}+frac{1}{4N-1}-(frac{1}{3}+frac{1}{7}+...frac{1}{4N-1})=sum_{n=1}^{2N}frac{1}{2n-1}-sum_{n=1}^{N}frac{1}{4n-1}$
Wich didn´t help because I find $sum_{n=1}^{N}frac{1}{4n-1}$ as difficult to solve as $sum_{n=1}^{N}frac{1}{4n-3}$ and I have no idea of what else to do.
sequences-and-series summation
edited Jan 28 at 16:53
Did
249k23226466
asked Jan 28 at 16:42
ChalklingChalkling
144
$endgroup$
We have been finding the value of sums related to the Harmonic series by using that
$sum_{n=1}^{N}frac{1}{n}= ln(N)+gamma + varepsilon_N$.
When I try to calculate $sum_{n=1}^{infty}frac{(-1)^{n}}{2n-1}$ using that method there is a point where I don´t know how to continue.
I have read that its value is $frac{pi}{4}$ but I don´t understand how is that obtained and I would like to know if it´s possible to solve it using the method I know.
This is what I did:
$S_N=sum_{n=1}^{N}frac{(-1)^n}{2n-1}$
$S_{2N}=sum_{n=1}^{2N}frac{(-1)^n}{2n-1}=-1+frac{1}{3}-frac{1}{5}+...-frac{1}{4N-3}+frac{1}{4N-1}=$
$=1+frac{1}{3}+frac{1}{5}+...+frac{1}{4N-1}-2(1+frac{1}{5}+frac{1}{9}+...+frac{1}{4N-3})=sum_{n=1}^{2N}frac{1}{2n-1}-2sum_{n=1}^{N}frac{1}{4n-3}$
.
$sum_{n=1}^{2N}frac{1}{2n-1}=1+frac{1}{3}+frac{1}{5}+...+frac{1}{4N-3}+frac{1}{4N-1}=$
$=1+frac{1}{2}+frac{1}{3}+...+frac{1}{4N-1}+frac{1}{4N}-(frac{1}{2}+frac{1}{4}+...+frac{1}{4N})=sum_{n=1}^{4N}frac{1}{n}-frac{1}{2}sum_{n=1}^{2N}frac{1}{n}$
.
But I don´t know what to do with the second one. I tried to write it like this:
$sum_{n=1}^{N}frac{1}{4n-3}=1+frac{1}{5}+frac{1}{9}+...+frac{1}{4N-3}=$
$=1+frac{1}{3}+frac{1}{5}+...+frac{1}{4N-3}+frac{1}{4N-1}-(frac{1}{3}+frac{1}{7}+...frac{1}{4N-1})=sum_{n=1}^{2N}frac{1}{2n-1}-sum_{n=1}^{N}frac{1}{4n-1}$
Wich didn´t help because I find $sum_{n=1}^{N}frac{1}{4n-1}$ as difficult to solve as $sum_{n=1}^{N}frac{1}{4n-3}$ and I have no idea of what else to do.
sequences-and-series summation
sequences-and-series summation
edited Jan 28 at 16:53
Did
249k23226466
asked Jan 28 at 16:42
ChalklingChalkling
144
edited Jan 28 at 16:53
Did
249k23226466
asked Jan 28 at 16:42
ChalklingChalkling
144
edited Jan 28 at 16:53
Did
249k23226466
edited Jan 28 at 16:53
Did
249k23226466
edited Jan 28 at 16:53
Did
249k23226466
249k23226466
asked Jan 28 at 16:42
ChalklingChalkling
144
asked Jan 28 at 16:42
ChalklingChalkling
144
asked Jan 28 at 16:42
ChalklingChalkling
144
144
1
$begingroup$
Hint: $$sum_{n=1}^{infty}frac{(-1)^{n}}{2n-1}=sum_{n=1}^{infty}frac{a_n}{n}$$ where $(a_n)$ has period $4$ hence $$a_n=sum_{k=1}^4alpha_ki^{kn}$$ since $i$ is a principal fourth root of $1$, for some given $(alpha_1,alpha_2,alpha_3,alpha_4)$ independent of $n$. Now, it happens that $alpha_4=0$. And remains to compute $$u(i^k)=sum_{n=1}^inftyfrac{i^{kn}}n$$ for $k=1$, $k=2$ and $k=3$. Can you do that? At the end, $$sum_{n=1}^{infty}frac{(-1)^{n}}{2n-1}=alpha_1u(i)+alpha_2u(-1)+alpha_3u(-i)$$
$endgroup$
– Did
Jan 28 at 16:49
$begingroup$
We are still doing only real analysis, so sadly I can´t use that
$endgroup$
– Chalkling
Jan 28 at 17:07
$begingroup$
The list of tools you cannot use grows longer any minute, it seems (see your comment to the answer below for another example). Say, what are the tools you can use?
$endgroup$
– Did
Jan 28 at 17:12
$begingroup$
Silence... Probably not either the identity $$picot(pi z)=frac1z+2zsum_{n=1}^inftyfrac1{z^2-n^2}$$ at $z=frac14$...
$endgroup$
– Did
Jan 28 at 17:23
$begingroup$
No, not either. As I said below, right now can´t use almost anything other than the method we have been taught to use for this series and I wanted to know if that was enogh
$endgroup$
– Chalkling
Jan 28 at 17:31
add a comment |
1
$begingroup$
Hint: $$sum_{n=1}^{infty}frac{(-1)^{n}}{2n-1}=sum_{n=1}^{infty}frac{a_n}{n}$$ where $(a_n)$ has period $4$ hence $$a_n=sum_{k=1}^4alpha_ki^{kn}$$ since $i$ is a principal fourth root of $1$, for some given $(alpha_1,alpha_2,alpha_3,alpha_4)$ independent of $n$. Now, it happens that $alpha_4=0$. And remains to compute $$u(i^k)=sum_{n=1}^inftyfrac{i^{kn}}n$$ for $k=1$, $k=2$ and $k=3$. Can you do that? At the end, $$sum_{n=1}^{infty}frac{(-1)^{n}}{2n-1}=alpha_1u(i)+alpha_2u(-1)+alpha_3u(-i)$$
$endgroup$
– Did
Jan 28 at 16:49
$begingroup$
We are still doing only real analysis, so sadly I can´t use that
$endgroup$
– Chalkling
Jan 28 at 17:07
$begingroup$
The list of tools you cannot use grows longer any minute, it seems (see your comment to the answer below for another example). Say, what are the tools you can use?
$endgroup$
– Did
Jan 28 at 17:12
$begingroup$
Silence... Probably not either the identity $$picot(pi z)=frac1z+2zsum_{n=1}^inftyfrac1{z^2-n^2}$$ at $z=frac14$...
$endgroup$
– Did
Jan 28 at 17:23
$begingroup$
No, not either. As I said below, right now can´t use almost anything other than the method we have been taught to use for this series and I wanted to know if that was enogh
$endgroup$
– Chalkling
Jan 28 at 17:31
1
1
$begingroup$
Hint: $$sum_{n=1}^{infty}frac{(-1)^{n}}{2n-1}=sum_{n=1}^{infty}frac{a_n}{n}$$ where $(a_n)$ has period $4$ hence $$a_n=sum_{k=1}^4alpha_ki^{kn}$$ since $i$ is a principal fourth root of $1$, for some given $(alpha_1,alpha_2,alpha_3,alpha_4)$ independent of $n$. Now, it happens that $alpha_4=0$. And remains to compute $$u(i^k)=sum_{n=1}^inftyfrac{i^{kn}}n$$ for $k=1$, $k=2$ and $k=3$. Can you do that? At the end, $$sum_{n=1}^{infty}frac{(-1)^{n}}{2n-1}=alpha_1u(i)+alpha_2u(-1)+alpha_3u(-i)$$
$endgroup$
– Did
Jan 28 at 16:49
$begingroup$
Hint: $$sum_{n=1}^{infty}frac{(-1)^{n}}{2n-1}=sum_{n=1}^{infty}frac{a_n}{n}$$ where $(a_n)$ has period $4$ hence $$a_n=sum_{k=1}^4alpha_ki^{kn}$$ since $i$ is a principal fourth root of $1$, for some given $(alpha_1,alpha_2,alpha_3,alpha_4)$ independent of $n$. Now, it happens that $alpha_4=0$. And remains to compute $$u(i^k)=sum_{n=1}^inftyfrac{i^{kn}}n$$ for $k=1$, $k=2$ and $k=3$. Can you do that? At the end, $$sum_{n=1}^{infty}frac{(-1)^{n}}{2n-1}=alpha_1u(i)+alpha_2u(-1)+alpha_3u(-i)$$
$endgroup$
– Did
Jan 28 at 16:49
$begingroup$
We are still doing only real analysis, so sadly I can´t use that
$endgroup$
– Chalkling
Jan 28 at 17:07
$begingroup$
We are still doing only real analysis, so sadly I can´t use that
$endgroup$
– Chalkling
Jan 28 at 17:07
$begingroup$
The list of tools you cannot use grows longer any minute, it seems (see your comment to the answer below for another example). Say, what are the tools you can use?
$endgroup$
– Did
Jan 28 at 17:12
$begingroup$
The list of tools you cannot use grows longer any minute, it seems (see your comment to the answer below for another example). Say, what are the tools you can use?
$endgroup$
– Did
Jan 28 at 17:12
$begingroup$
Silence... Probably not either the identity $$picot(pi z)=frac1z+2zsum_{n=1}^inftyfrac1{z^2-n^2}$$ at $z=frac14$...
$endgroup$
– Did
Jan 28 at 17:23
$begingroup$
Silence... Probably not either the identity $$picot(pi z)=frac1z+2zsum_{n=1}^inftyfrac1{z^2-n^2}$$ at $z=frac14$...
$endgroup$
– Did
Jan 28 at 17:23
$begingroup$
No, not either. As I said below, right now can´t use almost anything other than the method we have been taught to use for this series and I wanted to know if that was enogh
$endgroup$
– Chalkling
Jan 28 at 17:31
$begingroup$
No, not either. As I said below, right now can´t use almost anything other than the method we have been taught to use for this series and I wanted to know if that was enogh
$endgroup$
– Chalkling
Jan 28 at 17:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Indeed, this sum requires more than playing with the asymptotic development of the harmonic series.
I suggest that you consider “instead” $$S_n=sum_{k=0}^n{frac{(-1)^n}{2n+1}}=int_0^1{sum_{k=0}^n{(-x^2)^n},dx},$$ and then prove that $$left|S_n-int_0^1{frac{1}{1+x^2},dx}right| leq n^{-1}.$$
edited Jan 28 at 16:51
gt6989b
35.2k22557
answered Jan 28 at 16:49
MindlackMindlack
4,910211
$endgroup$
$begingroup$
Problem is, we haven´t learned integration yet
$endgroup$
– Chalkling
Jan 28 at 17:04
$begingroup$
Is this homework? What are the mathematical tools you can/may/know how to use?
$endgroup$
– Mindlack
Jan 28 at 17:09
$begingroup$
Well, its not homework, just something I found while I was practising similar series. My professor looked at it briefly and said we could probably solve it, but we have started with series recently and thats pretty much the only way I know to do this type. I wanted to know if I was missing something that allowed me to do it like that.
$endgroup$
– Chalkling
Jan 28 at 17:24
$begingroup$
Ok. But I do not know of any way to involve $pi$ in a calculation that uses neither integrals nor a bit of complex analysis. Besides, if your partial sum could be expressed with harmonic sums, you would get a relation between $gamma$ and $pi$ (The rest of the expansion has rational coefficients), and I am quite sure we do not really know one.
$endgroup$
– Mindlack
Jan 28 at 17:54
add a comment |
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1 Answer
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$begingroup$
Indeed, this sum requires more than playing with the asymptotic development of the harmonic series.
I suggest that you consider “instead” $$S_n=sum_{k=0}^n{frac{(-1)^n}{2n+1}}=int_0^1{sum_{k=0}^n{(-x^2)^n},dx},$$ and then prove that $$left|S_n-int_0^1{frac{1}{1+x^2},dx}right| leq n^{-1}.$$
edited Jan 28 at 16:51
gt6989b
35.2k22557
answered Jan 28 at 16:49
MindlackMindlack
4,910211
$endgroup$
$begingroup$
Problem is, we haven´t learned integration yet
$endgroup$
– Chalkling
Jan 28 at 17:04
$begingroup$
Is this homework? What are the mathematical tools you can/may/know how to use?
$endgroup$
– Mindlack
Jan 28 at 17:09
$begingroup$
Well, its not homework, just something I found while I was practising similar series. My professor looked at it briefly and said we could probably solve it, but we have started with series recently and thats pretty much the only way I know to do this type. I wanted to know if I was missing something that allowed me to do it like that.
$endgroup$
– Chalkling
Jan 28 at 17:24
$begingroup$
Ok. But I do not know of any way to involve $pi$ in a calculation that uses neither integrals nor a bit of complex analysis. Besides, if your partial sum could be expressed with harmonic sums, you would get a relation between $gamma$ and $pi$ (The rest of the expansion has rational coefficients), and I am quite sure we do not really know one.
$endgroup$
– Mindlack
Jan 28 at 17:54
add a comment |
$begingroup$
Indeed, this sum requires more than playing with the asymptotic development of the harmonic series.
I suggest that you consider “instead” $$S_n=sum_{k=0}^n{frac{(-1)^n}{2n+1}}=int_0^1{sum_{k=0}^n{(-x^2)^n},dx},$$ and then prove that $$left|S_n-int_0^1{frac{1}{1+x^2},dx}right| leq n^{-1}.$$
edited Jan 28 at 16:51
gt6989b
35.2k22557
answered Jan 28 at 16:49
MindlackMindlack
4,910211
$endgroup$
$begingroup$
Problem is, we haven´t learned integration yet
$endgroup$
– Chalkling
Jan 28 at 17:04
$begingroup$
Is this homework? What are the mathematical tools you can/may/know how to use?
$endgroup$
– Mindlack
Jan 28 at 17:09
$begingroup$
Well, its not homework, just something I found while I was practising similar series. My professor looked at it briefly and said we could probably solve it, but we have started with series recently and thats pretty much the only way I know to do this type. I wanted to know if I was missing something that allowed me to do it like that.
$endgroup$
– Chalkling
Jan 28 at 17:24
$begingroup$
Ok. But I do not know of any way to involve $pi$ in a calculation that uses neither integrals nor a bit of complex analysis. Besides, if your partial sum could be expressed with harmonic sums, you would get a relation between $gamma$ and $pi$ (The rest of the expansion has rational coefficients), and I am quite sure we do not really know one.
$endgroup$
– Mindlack
Jan 28 at 17:54
add a comment |
$begingroup$
Indeed, this sum requires more than playing with the asymptotic development of the harmonic series.
I suggest that you consider “instead” $$S_n=sum_{k=0}^n{frac{(-1)^n}{2n+1}}=int_0^1{sum_{k=0}^n{(-x^2)^n},dx},$$ and then prove that $$left|S_n-int_0^1{frac{1}{1+x^2},dx}right| leq n^{-1}.$$
edited Jan 28 at 16:51
gt6989b
35.2k22557
answered Jan 28 at 16:49
MindlackMindlack
4,910211
$endgroup$
Indeed, this sum requires more than playing with the asymptotic development of the harmonic series.
I suggest that you consider “instead” $$S_n=sum_{k=0}^n{frac{(-1)^n}{2n+1}}=int_0^1{sum_{k=0}^n{(-x^2)^n},dx},$$ and then prove that $$left|S_n-int_0^1{frac{1}{1+x^2},dx}right| leq n^{-1}.$$
edited Jan 28 at 16:51
gt6989b
35.2k22557
answered Jan 28 at 16:49
MindlackMindlack
4,910211
edited Jan 28 at 16:51
gt6989b
35.2k22557
edited Jan 28 at 16:51
gt6989b
35.2k22557
edited Jan 28 at 16:51
gt6989b
35.2k22557
35.2k22557
answered Jan 28 at 16:49
MindlackMindlack
4,910211
answered Jan 28 at 16:49
MindlackMindlack
4,910211
answered Jan 28 at 16:49
MindlackMindlack
4,910211
4,910211
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Problem is, we haven´t learned integration yet
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– Chalkling
Jan 28 at 17:04
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Is this homework? What are the mathematical tools you can/may/know how to use?
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– Mindlack
Jan 28 at 17:09
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Well, its not homework, just something I found while I was practising similar series. My professor looked at it briefly and said we could probably solve it, but we have started with series recently and thats pretty much the only way I know to do this type. I wanted to know if I was missing something that allowed me to do it like that.
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– Chalkling
Jan 28 at 17:24
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Ok. But I do not know of any way to involve $pi$ in a calculation that uses neither integrals nor a bit of complex analysis. Besides, if your partial sum could be expressed with harmonic sums, you would get a relation between $gamma$ and $pi$ (The rest of the expansion has rational coefficients), and I am quite sure we do not really know one.
$endgroup$
– Mindlack
Jan 28 at 17:54
add a comment |
$begingroup$
Problem is, we haven´t learned integration yet
$endgroup$
– Chalkling
Jan 28 at 17:04
$begingroup$
Is this homework? What are the mathematical tools you can/may/know how to use?
$endgroup$
– Mindlack
Jan 28 at 17:09
$begingroup$
Well, its not homework, just something I found while I was practising similar series. My professor looked at it briefly and said we could probably solve it, but we have started with series recently and thats pretty much the only way I know to do this type. I wanted to know if I was missing something that allowed me to do it like that.
$endgroup$
– Chalkling
Jan 28 at 17:24
$begingroup$
Ok. But I do not know of any way to involve $pi$ in a calculation that uses neither integrals nor a bit of complex analysis. Besides, if your partial sum could be expressed with harmonic sums, you would get a relation between $gamma$ and $pi$ (The rest of the expansion has rational coefficients), and I am quite sure we do not really know one.
$endgroup$
– Mindlack
Jan 28 at 17:54
$begingroup$
Problem is, we haven´t learned integration yet
$endgroup$
– Chalkling
Jan 28 at 17:04
$begingroup$
Problem is, we haven´t learned integration yet
$endgroup$
– Chalkling
Jan 28 at 17:04
$begingroup$
Is this homework? What are the mathematical tools you can/may/know how to use?
$endgroup$
– Mindlack
Jan 28 at 17:09
$begingroup$
Is this homework? What are the mathematical tools you can/may/know how to use?
$endgroup$
– Mindlack
Jan 28 at 17:09
$begingroup$
Well, its not homework, just something I found while I was practising similar series. My professor looked at it briefly and said we could probably solve it, but we have started with series recently and thats pretty much the only way I know to do this type. I wanted to know if I was missing something that allowed me to do it like that.
$endgroup$
– Chalkling
Jan 28 at 17:24
$begingroup$
Well, its not homework, just something I found while I was practising similar series. My professor looked at it briefly and said we could probably solve it, but we have started with series recently and thats pretty much the only way I know to do this type. I wanted to know if I was missing something that allowed me to do it like that.
$endgroup$
– Chalkling
Jan 28 at 17:24
$begingroup$
Ok. But I do not know of any way to involve $pi$ in a calculation that uses neither integrals nor a bit of complex analysis. Besides, if your partial sum could be expressed with harmonic sums, you would get a relation between $gamma$ and $pi$ (The rest of the expansion has rational coefficients), and I am quite sure we do not really know one.
$endgroup$
– Mindlack
Jan 28 at 17:54
$begingroup$
Ok. But I do not know of any way to involve $pi$ in a calculation that uses neither integrals nor a bit of complex analysis. Besides, if your partial sum could be expressed with harmonic sums, you would get a relation between $gamma$ and $pi$ (The rest of the expansion has rational coefficients), and I am quite sure we do not really know one.
$endgroup$
– Mindlack
Jan 28 at 17:54
add a comment |
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Hint: $$sum_{n=1}^{infty}frac{(-1)^{n}}{2n-1}=sum_{n=1}^{infty}frac{a_n}{n}$$ where $(a_n)$ has period $4$ hence $$a_n=sum_{k=1}^4alpha_ki^{kn}$$ since $i$ is a principal fourth root of $1$, for some given $(alpha_1,alpha_2,alpha_3,alpha_4)$ independent of $n$. Now, it happens that $alpha_4=0$. And remains to compute $$u(i^k)=sum_{n=1}^inftyfrac{i^{kn}}n$$ for $k=1$, $k=2$ and $k=3$. Can you do that? At the end, $$sum_{n=1}^{infty}frac{(-1)^{n}}{2n-1}=alpha_1u(i)+alpha_2u(-1)+alpha_3u(-i)$$
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– Did
Jan 28 at 16:49
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We are still doing only real analysis, so sadly I can´t use that
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– Chalkling
Jan 28 at 17:07
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The list of tools you cannot use grows longer any minute, it seems (see your comment to the answer below for another example). Say, what are the tools you can use?
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– Did
Jan 28 at 17:12
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Silence... Probably not either the identity $$picot(pi z)=frac1z+2zsum_{n=1}^inftyfrac1{z^2-n^2}$$ at $z=frac14$...
$endgroup$
– Did
Jan 28 at 17:23
$begingroup$
No, not either. As I said below, right now can´t use almost anything other than the method we have been taught to use for this series and I wanted to know if that was enogh
$endgroup$
– Chalkling
Jan 28 at 17:31