Mixing
Does $(1+frac12-frac13) + (frac14+frac15-frac16)+(frac17+frac18-frac19)+cdots$ converge?
$begingroup$
Does the series $$S=left(1+frac{1}{2}-frac{1}{3} right) + left(frac{1}{4}+frac{1}{5}-frac{1}{6} right)+left(frac{1}{7}+frac{1}{8}-frac{1}{9}right)+cdots$$ converge?
Here's my attempt at a solution: $$S = sum_{n=1}^{infty}frac{1}{n}-2sum_{n=1}^{infty}frac{1}{3n}=sum_{n=1}^{infty}frac{1}{3n}=frac{1}{3}sum_{n=1}^{infty}frac{1}{n}$$
As we can "rewrite" this series as one third of the harmonical series (that diverges), we conclude the divergence of $S$.
Is this right? Which other convergence tests could be used?
calculus sequences-and-series proof-verification convergence divergent-series
edited Jan 6 at 18:08
Did
247k23223459
asked Jan 6 at 2:43
Raúl AsteteRaúl Astete
516
$endgroup$
add a comment |
$begingroup$
Does the series $$S=left(1+frac{1}{2}-frac{1}{3} right) + left(frac{1}{4}+frac{1}{5}-frac{1}{6} right)+left(frac{1}{7}+frac{1}{8}-frac{1}{9}right)+cdots$$ converge?
Here's my attempt at a solution: $$S = sum_{n=1}^{infty}frac{1}{n}-2sum_{n=1}^{infty}frac{1}{3n}=sum_{n=1}^{infty}frac{1}{3n}=frac{1}{3}sum_{n=1}^{infty}frac{1}{n}$$
As we can "rewrite" this series as one third of the harmonical series (that diverges), we conclude the divergence of $S$.
Is this right? Which other convergence tests could be used?
calculus sequences-and-series proof-verification convergence divergent-series
edited Jan 6 at 18:08
Did
247k23223459
asked Jan 6 at 2:43
Raúl AsteteRaúl Astete
516
$endgroup$
2
$begingroup$
your solution is not correct. You cannot change the order of summation of a series that doesn't converge absolutely, neither make (standard) arithmetic with divergent series
$endgroup$
– Masacroso
Jan 6 at 3:15
$begingroup$
@Masacroso Is it not fine to do something along the lines of " Suppose the following series converges. Then in particular, it converges absolutely, so we can reorder the series in any way. One particular ordering gives a scalar multiple of the harmonic series which diverges, so the given series cannot converge".
$endgroup$
– Saad
Jan 6 at 3:21
5
$begingroup$
@Saad That is not a valid argument because convergence doesn't imply absolute convergence (you've got it the wrong way round).
$endgroup$
– AlephNull
Jan 6 at 12:29
$begingroup$
Of course, you're correct. For whatever reason, I thought each term was positive..
$endgroup$
– Saad
Jan 7 at 15:19
add a comment |
$begingroup$
Does the series $$S=left(1+frac{1}{2}-frac{1}{3} right) + left(frac{1}{4}+frac{1}{5}-frac{1}{6} right)+left(frac{1}{7}+frac{1}{8}-frac{1}{9}right)+cdots$$ converge?
Here's my attempt at a solution: $$S = sum_{n=1}^{infty}frac{1}{n}-2sum_{n=1}^{infty}frac{1}{3n}=sum_{n=1}^{infty}frac{1}{3n}=frac{1}{3}sum_{n=1}^{infty}frac{1}{n}$$
As we can "rewrite" this series as one third of the harmonical series (that diverges), we conclude the divergence of $S$.
Is this right? Which other convergence tests could be used?
calculus sequences-and-series proof-verification convergence divergent-series
edited Jan 6 at 18:08
Did
247k23223459
asked Jan 6 at 2:43
Raúl AsteteRaúl Astete
516
$endgroup$
Does the series $$S=left(1+frac{1}{2}-frac{1}{3} right) + left(frac{1}{4}+frac{1}{5}-frac{1}{6} right)+left(frac{1}{7}+frac{1}{8}-frac{1}{9}right)+cdots$$ converge?
Here's my attempt at a solution: $$S = sum_{n=1}^{infty}frac{1}{n}-2sum_{n=1}^{infty}frac{1}{3n}=sum_{n=1}^{infty}frac{1}{3n}=frac{1}{3}sum_{n=1}^{infty}frac{1}{n}$$
As we can "rewrite" this series as one third of the harmonical series (that diverges), we conclude the divergence of $S$.
Is this right? Which other convergence tests could be used?
calculus sequences-and-series proof-verification convergence divergent-series
calculus sequences-and-series proof-verification convergence divergent-series
edited Jan 6 at 18:08
Did
247k23223459
asked Jan 6 at 2:43
Raúl AsteteRaúl Astete
516
edited Jan 6 at 18:08
Did
247k23223459
asked Jan 6 at 2:43
Raúl AsteteRaúl Astete
516
edited Jan 6 at 18:08
Did
247k23223459
edited Jan 6 at 18:08
Did
247k23223459
edited Jan 6 at 18:08
Did
247k23223459
247k23223459
asked Jan 6 at 2:43
Raúl AsteteRaúl Astete
516
asked Jan 6 at 2:43
Raúl AsteteRaúl Astete
516
asked Jan 6 at 2:43
Raúl AsteteRaúl Astete
516
516
2
$begingroup$
your solution is not correct. You cannot change the order of summation of a series that doesn't converge absolutely, neither make (standard) arithmetic with divergent series
$endgroup$
– Masacroso
Jan 6 at 3:15
$begingroup$
@Masacroso Is it not fine to do something along the lines of " Suppose the following series converges. Then in particular, it converges absolutely, so we can reorder the series in any way. One particular ordering gives a scalar multiple of the harmonic series which diverges, so the given series cannot converge".
$endgroup$
– Saad
Jan 6 at 3:21
5
$begingroup$
@Saad That is not a valid argument because convergence doesn't imply absolute convergence (you've got it the wrong way round).
$endgroup$
– AlephNull
Jan 6 at 12:29
$begingroup$
Of course, you're correct. For whatever reason, I thought each term was positive..
$endgroup$
– Saad
Jan 7 at 15:19
add a comment |
2
$begingroup$
your solution is not correct. You cannot change the order of summation of a series that doesn't converge absolutely, neither make (standard) arithmetic with divergent series
$endgroup$
– Masacroso
Jan 6 at 3:15
$begingroup$
@Masacroso Is it not fine to do something along the lines of " Suppose the following series converges. Then in particular, it converges absolutely, so we can reorder the series in any way. One particular ordering gives a scalar multiple of the harmonic series which diverges, so the given series cannot converge".
$endgroup$
– Saad
Jan 6 at 3:21
5
$begingroup$
@Saad That is not a valid argument because convergence doesn't imply absolute convergence (you've got it the wrong way round).
$endgroup$
– AlephNull
Jan 6 at 12:29
$begingroup$
Of course, you're correct. For whatever reason, I thought each term was positive..
$endgroup$
– Saad
Jan 7 at 15:19
2
2
$begingroup$
your solution is not correct. You cannot change the order of summation of a series that doesn't converge absolutely, neither make (standard) arithmetic with divergent series
$endgroup$
– Masacroso
Jan 6 at 3:15
$begingroup$
your solution is not correct. You cannot change the order of summation of a series that doesn't converge absolutely, neither make (standard) arithmetic with divergent series
$endgroup$
– Masacroso
Jan 6 at 3:15
$begingroup$
@Masacroso Is it not fine to do something along the lines of " Suppose the following series converges. Then in particular, it converges absolutely, so we can reorder the series in any way. One particular ordering gives a scalar multiple of the harmonic series which diverges, so the given series cannot converge".
$endgroup$
– Saad
Jan 6 at 3:21
$begingroup$
@Masacroso Is it not fine to do something along the lines of " Suppose the following series converges. Then in particular, it converges absolutely, so we can reorder the series in any way. One particular ordering gives a scalar multiple of the harmonic series which diverges, so the given series cannot converge".
$endgroup$
– Saad
Jan 6 at 3:21
5
5
$begingroup$
@Saad That is not a valid argument because convergence doesn't imply absolute convergence (you've got it the wrong way round).
$endgroup$
– AlephNull
Jan 6 at 12:29
$begingroup$
@Saad That is not a valid argument because convergence doesn't imply absolute convergence (you've got it the wrong way round).
$endgroup$
– AlephNull
Jan 6 at 12:29
$begingroup$
Of course, you're correct. For whatever reason, I thought each term was positive..
$endgroup$
– Saad
Jan 7 at 15:19
$begingroup$
Of course, you're correct. For whatever reason, I thought each term was positive..
$endgroup$
– Saad
Jan 7 at 15:19
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your answer is correct, but your reasoning is not. Order matters. You should write the sum as follows:
$$sum_{n=0}^inftyleft(frac{1}{3n+1}+frac{1}{3n+2}-frac{1}{3n+3}right)$$
Simplify what is in the parentheses and then evaluate in the usual way.
Oh, and it is true that order does not matter if all the terms are positive. But when some terms are positive and others negative, you have to be more careful.
answered Jan 6 at 3:03
Ben WBen W
2,234615
$endgroup$
2
$begingroup$
In this case you can simply say $frac 1{3n+2}-frac 1{3n+3}>0$ thus the series is $>sumfrac 1{3n+1}$ which is a divergent series. I mean we do not even need to calculate that the parenthesis is equivalent to $frac {9n^2}{27n^3}$.
$endgroup$
– zwim
Jan 6 at 4:32
$begingroup$
@zwim quite true, that is a nice little shortcut
$endgroup$
– Ben W
Jan 6 at 14:47
$begingroup$
Simplify and it behaves like $1/n$, so it diverges.
$endgroup$
– ncmathsadist
Jan 6 at 18:12
add a comment |
$begingroup$
The idea is correct, but you cannot write like that, because the series involved are both divergent and generally it is moot to do arithmetic operations of divergent series.
Let $H_n = sum_1^n 1/k$, i.e. the $n$-th partial sum of the harmonic series, then the
$$
S_n = H_{3n} - frac 23 H_n,
$$
then use the asymptotic expression $H_n = log n + gamma + varepsilon_n$ where $varepsilon_n to 0 [n to infty]$, we have
$$
S_n = log (3n) -frac 23log n + frac 13 gamma + varepsilon_{3n} - frac 23 varepsilon_n = log(3n^{1/3}) + frac 13 gamma + alpha_n xrightarrow{n to infty} +infty.
$$
answered Jan 6 at 3:13
xbhxbh
6,1051522
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your answer is correct, but your reasoning is not. Order matters. You should write the sum as follows:
$$sum_{n=0}^inftyleft(frac{1}{3n+1}+frac{1}{3n+2}-frac{1}{3n+3}right)$$
Simplify what is in the parentheses and then evaluate in the usual way.
Oh, and it is true that order does not matter if all the terms are positive. But when some terms are positive and others negative, you have to be more careful.
answered Jan 6 at 3:03
Ben WBen W
2,234615
$endgroup$
2
$begingroup$
In this case you can simply say $frac 1{3n+2}-frac 1{3n+3}>0$ thus the series is $>sumfrac 1{3n+1}$ which is a divergent series. I mean we do not even need to calculate that the parenthesis is equivalent to $frac {9n^2}{27n^3}$.
$endgroup$
– zwim
Jan 6 at 4:32
$begingroup$
@zwim quite true, that is a nice little shortcut
$endgroup$
– Ben W
Jan 6 at 14:47
$begingroup$
Simplify and it behaves like $1/n$, so it diverges.
$endgroup$
– ncmathsadist
Jan 6 at 18:12
add a comment |
$begingroup$
Your answer is correct, but your reasoning is not. Order matters. You should write the sum as follows:
$$sum_{n=0}^inftyleft(frac{1}{3n+1}+frac{1}{3n+2}-frac{1}{3n+3}right)$$
Simplify what is in the parentheses and then evaluate in the usual way.
Oh, and it is true that order does not matter if all the terms are positive. But when some terms are positive and others negative, you have to be more careful.
answered Jan 6 at 3:03
Ben WBen W
2,234615
$endgroup$
2
$begingroup$
In this case you can simply say $frac 1{3n+2}-frac 1{3n+3}>0$ thus the series is $>sumfrac 1{3n+1}$ which is a divergent series. I mean we do not even need to calculate that the parenthesis is equivalent to $frac {9n^2}{27n^3}$.
$endgroup$
– zwim
Jan 6 at 4:32
$begingroup$
@zwim quite true, that is a nice little shortcut
$endgroup$
– Ben W
Jan 6 at 14:47
$begingroup$
Simplify and it behaves like $1/n$, so it diverges.
$endgroup$
– ncmathsadist
Jan 6 at 18:12
add a comment |
$begingroup$
Your answer is correct, but your reasoning is not. Order matters. You should write the sum as follows:
$$sum_{n=0}^inftyleft(frac{1}{3n+1}+frac{1}{3n+2}-frac{1}{3n+3}right)$$
Simplify what is in the parentheses and then evaluate in the usual way.
Oh, and it is true that order does not matter if all the terms are positive. But when some terms are positive and others negative, you have to be more careful.
answered Jan 6 at 3:03
Ben WBen W
2,234615
$endgroup$
Your answer is correct, but your reasoning is not. Order matters. You should write the sum as follows:
$$sum_{n=0}^inftyleft(frac{1}{3n+1}+frac{1}{3n+2}-frac{1}{3n+3}right)$$
Simplify what is in the parentheses and then evaluate in the usual way.
Oh, and it is true that order does not matter if all the terms are positive. But when some terms are positive and others negative, you have to be more careful.
answered Jan 6 at 3:03
Ben WBen W
2,234615
edited Jan 6 at 3:09
answered Jan 6 at 3:03
Ben WBen W
2,234615
answered Jan 6 at 3:03
Ben WBen W
2,234615
answered Jan 6 at 3:03
Ben WBen W
2,234615
2,234615
2
$begingroup$
In this case you can simply say $frac 1{3n+2}-frac 1{3n+3}>0$ thus the series is $>sumfrac 1{3n+1}$ which is a divergent series. I mean we do not even need to calculate that the parenthesis is equivalent to $frac {9n^2}{27n^3}$.
$endgroup$
– zwim
Jan 6 at 4:32
$begingroup$
@zwim quite true, that is a nice little shortcut
$endgroup$
– Ben W
Jan 6 at 14:47
$begingroup$
Simplify and it behaves like $1/n$, so it diverges.
$endgroup$
– ncmathsadist
Jan 6 at 18:12
add a comment |
2
$begingroup$
In this case you can simply say $frac 1{3n+2}-frac 1{3n+3}>0$ thus the series is $>sumfrac 1{3n+1}$ which is a divergent series. I mean we do not even need to calculate that the parenthesis is equivalent to $frac {9n^2}{27n^3}$.
$endgroup$
– zwim
Jan 6 at 4:32
$begingroup$
@zwim quite true, that is a nice little shortcut
$endgroup$
– Ben W
Jan 6 at 14:47
$begingroup$
Simplify and it behaves like $1/n$, so it diverges.
$endgroup$
– ncmathsadist
Jan 6 at 18:12
2
2
$begingroup$
In this case you can simply say $frac 1{3n+2}-frac 1{3n+3}>0$ thus the series is $>sumfrac 1{3n+1}$ which is a divergent series. I mean we do not even need to calculate that the parenthesis is equivalent to $frac {9n^2}{27n^3}$.
$endgroup$
– zwim
Jan 6 at 4:32
$begingroup$
In this case you can simply say $frac 1{3n+2}-frac 1{3n+3}>0$ thus the series is $>sumfrac 1{3n+1}$ which is a divergent series. I mean we do not even need to calculate that the parenthesis is equivalent to $frac {9n^2}{27n^3}$.
$endgroup$
– zwim
Jan 6 at 4:32
$begingroup$
@zwim quite true, that is a nice little shortcut
$endgroup$
– Ben W
Jan 6 at 14:47
$begingroup$
@zwim quite true, that is a nice little shortcut
$endgroup$
– Ben W
Jan 6 at 14:47
$begingroup$
Simplify and it behaves like $1/n$, so it diverges.
$endgroup$
– ncmathsadist
Jan 6 at 18:12
$begingroup$
Simplify and it behaves like $1/n$, so it diverges.
$endgroup$
– ncmathsadist
Jan 6 at 18:12
add a comment |
$begingroup$
The idea is correct, but you cannot write like that, because the series involved are both divergent and generally it is moot to do arithmetic operations of divergent series.
Let $H_n = sum_1^n 1/k$, i.e. the $n$-th partial sum of the harmonic series, then the
$$
S_n = H_{3n} - frac 23 H_n,
$$
then use the asymptotic expression $H_n = log n + gamma + varepsilon_n$ where $varepsilon_n to 0 [n to infty]$, we have
$$
S_n = log (3n) -frac 23log n + frac 13 gamma + varepsilon_{3n} - frac 23 varepsilon_n = log(3n^{1/3}) + frac 13 gamma + alpha_n xrightarrow{n to infty} +infty.
$$
answered Jan 6 at 3:13
xbhxbh
6,1051522
$endgroup$
add a comment |
$begingroup$
The idea is correct, but you cannot write like that, because the series involved are both divergent and generally it is moot to do arithmetic operations of divergent series.
Let $H_n = sum_1^n 1/k$, i.e. the $n$-th partial sum of the harmonic series, then the
$$
S_n = H_{3n} - frac 23 H_n,
$$
then use the asymptotic expression $H_n = log n + gamma + varepsilon_n$ where $varepsilon_n to 0 [n to infty]$, we have
$$
S_n = log (3n) -frac 23log n + frac 13 gamma + varepsilon_{3n} - frac 23 varepsilon_n = log(3n^{1/3}) + frac 13 gamma + alpha_n xrightarrow{n to infty} +infty.
$$
answered Jan 6 at 3:13
xbhxbh
6,1051522
$endgroup$
add a comment |
$begingroup$
The idea is correct, but you cannot write like that, because the series involved are both divergent and generally it is moot to do arithmetic operations of divergent series.
Let $H_n = sum_1^n 1/k$, i.e. the $n$-th partial sum of the harmonic series, then the
$$
S_n = H_{3n} - frac 23 H_n,
$$
then use the asymptotic expression $H_n = log n + gamma + varepsilon_n$ where $varepsilon_n to 0 [n to infty]$, we have
$$
S_n = log (3n) -frac 23log n + frac 13 gamma + varepsilon_{3n} - frac 23 varepsilon_n = log(3n^{1/3}) + frac 13 gamma + alpha_n xrightarrow{n to infty} +infty.
$$
answered Jan 6 at 3:13
xbhxbh
6,1051522
$endgroup$
The idea is correct, but you cannot write like that, because the series involved are both divergent and generally it is moot to do arithmetic operations of divergent series.
Let $H_n = sum_1^n 1/k$, i.e. the $n$-th partial sum of the harmonic series, then the
$$
S_n = H_{3n} - frac 23 H_n,
$$
then use the asymptotic expression $H_n = log n + gamma + varepsilon_n$ where $varepsilon_n to 0 [n to infty]$, we have
$$
S_n = log (3n) -frac 23log n + frac 13 gamma + varepsilon_{3n} - frac 23 varepsilon_n = log(3n^{1/3}) + frac 13 gamma + alpha_n xrightarrow{n to infty} +infty.
$$
answered Jan 6 at 3:13
xbhxbh
6,1051522
answered Jan 6 at 3:13
xbhxbh
6,1051522
answered Jan 6 at 3:13
xbhxbh
6,1051522
answered Jan 6 at 3:13
xbhxbh
6,1051522
6,1051522
add a comment |
add a comment |
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2
$begingroup$
your solution is not correct. You cannot change the order of summation of a series that doesn't converge absolutely, neither make (standard) arithmetic with divergent series
$endgroup$
– Masacroso
Jan 6 at 3:15
$begingroup$
@Masacroso Is it not fine to do something along the lines of " Suppose the following series converges. Then in particular, it converges absolutely, so we can reorder the series in any way. One particular ordering gives a scalar multiple of the harmonic series which diverges, so the given series cannot converge".
$endgroup$
– Saad
Jan 6 at 3:21
5
$begingroup$
@Saad That is not a valid argument because convergence doesn't imply absolute convergence (you've got it the wrong way round).
$endgroup$
– AlephNull
Jan 6 at 12:29
$begingroup$
Of course, you're correct. For whatever reason, I thought each term was positive..
$endgroup$
– Saad
Jan 7 at 15:19