Mixing
Why does this sum $sumlimits_{n=1}^{infty}{frac{2^{2^{n-1}}}{2^{2^n}-1}}$ converge to 1?
$begingroup$
A friend of mine gave me this quiz: Where does this sum converge to?
$$sumlimits_{n=1}^{infty}{frac{2^{2^{n-1}}}{2^{2^n}-1}}$$
$a_1=frac{2}{3} , a_2=frac{4}{15}, a_3=frac{16}{255},...$and $S_1=frac{2}{3}, S_2=frac{14}{15} , S_3=frac{254}{255}, ...$
So I thought that this sum will converge to 1.
Let $$2^{2^{n-1}}=A_n, a_n=frac{A_n}{A_n^2-1}=frac{1}{2}left(frac{1}{A_n+1}+frac{1}{A_n-1}right)$$
Since $A_n = A_{n-1}^2$, I thought that this could lead to finding the partial sum of the series. I could prove that $$S_n=1-frac{1}{A_n^2-1}$$by mathematical induction, but I want to know whether my method can be useful for finding $S_n$ from scratch.
So my question is : Is there a way to find $S_n$ from scratch??
Also, I found that $$a_n=frac{frac{1}{A_n}}{1-frac{1}{A_n^2}}=sum_{n=1}^{infty}{frac{1}{A_n^{2n-1}}}=frac{1}{2^{1cdot2^{n-1}}}+frac{1}{2^{3cdot 2^{n-1}}}+frac{1}{2^{5cdot 2^{n-1}}}+....\
therefore sum_{n=1}^infty {a_n}=frac{1}{2^{1cdot 2^{0}}}+frac{1}{2^{3cdot 2^{0}}}+frac{1}{2^{5cdot 2^{0}}}+...\
+frac{1}{2^{1cdot 2^{1}}}+frac{1}{2^{3cdot 2^{1}}}+frac{1}{2^{5cdot 2^{1}}}+...\ +frac{1}{2^{1cdot 2^{2}}}+frac{1}{2^{3cdot 2^{2}}}+frac{1}{2^{5cdot 2^{2}}}+...
\ =sum_{n=1}^{infty}{2^{-n}}=1$$
Is this a valid way? I used the Riemann series theorem to rearrange them.
sequences-and-series
asked Jan 17 at 14:18
John. PJohn. P
1789
$endgroup$
|
show 1 more comment
$begingroup$
A friend of mine gave me this quiz: Where does this sum converge to?
$$sumlimits_{n=1}^{infty}{frac{2^{2^{n-1}}}{2^{2^n}-1}}$$
$a_1=frac{2}{3} , a_2=frac{4}{15}, a_3=frac{16}{255},...$and $S_1=frac{2}{3}, S_2=frac{14}{15} , S_3=frac{254}{255}, ...$
So I thought that this sum will converge to 1.
Let $$2^{2^{n-1}}=A_n, a_n=frac{A_n}{A_n^2-1}=frac{1}{2}left(frac{1}{A_n+1}+frac{1}{A_n-1}right)$$
Since $A_n = A_{n-1}^2$, I thought that this could lead to finding the partial sum of the series. I could prove that $$S_n=1-frac{1}{A_n^2-1}$$by mathematical induction, but I want to know whether my method can be useful for finding $S_n$ from scratch.
So my question is : Is there a way to find $S_n$ from scratch??
Also, I found that $$a_n=frac{frac{1}{A_n}}{1-frac{1}{A_n^2}}=sum_{n=1}^{infty}{frac{1}{A_n^{2n-1}}}=frac{1}{2^{1cdot2^{n-1}}}+frac{1}{2^{3cdot 2^{n-1}}}+frac{1}{2^{5cdot 2^{n-1}}}+....\
therefore sum_{n=1}^infty {a_n}=frac{1}{2^{1cdot 2^{0}}}+frac{1}{2^{3cdot 2^{0}}}+frac{1}{2^{5cdot 2^{0}}}+...\
+frac{1}{2^{1cdot 2^{1}}}+frac{1}{2^{3cdot 2^{1}}}+frac{1}{2^{5cdot 2^{1}}}+...\ +frac{1}{2^{1cdot 2^{2}}}+frac{1}{2^{3cdot 2^{2}}}+frac{1}{2^{5cdot 2^{2}}}+...
\ =sum_{n=1}^{infty}{2^{-n}}=1$$
Is this a valid way? I used the Riemann series theorem to rearrange them.
sequences-and-series
asked Jan 17 at 14:18
John. PJohn. P
1789
$endgroup$
1
$begingroup$
This is valid, since the elements are non-negative,
$endgroup$
– Bonbon
Jan 17 at 14:23
$begingroup$
@Bonbon Thx for replying!!
$endgroup$
– John. P
Jan 17 at 15:24
$begingroup$
It is also interesting to show that $$ 1 = sum_{kgeq 0}frac{2^k}{2^{2^k}+1}.$$
$endgroup$
– Jack D'Aurizio
Jan 17 at 19:19
$begingroup$
Have a look at this and replace $z=frac{1}{2}$.
$endgroup$
– rtybase
Jan 17 at 21:41
$begingroup$
Notice that $frac{2^{2^{n-1}}}{2^{2^n}-1} = frac{2^{2^{n-1}}}{(2^{2^{n-1}}-1)(2^{2^{n-1}}+1)} = frac{1}{2^{2^{n-1}}-1}-frac{1}{2^{2^n}-1} equiv b_{n-1} - b_n$ where $b_n = frac{1}{2^{2^n}-1}$. Telescoping!
$endgroup$
– Winther
Jan 17 at 23:49
|
show 1 more comment
$begingroup$
A friend of mine gave me this quiz: Where does this sum converge to?
$$sumlimits_{n=1}^{infty}{frac{2^{2^{n-1}}}{2^{2^n}-1}}$$
$a_1=frac{2}{3} , a_2=frac{4}{15}, a_3=frac{16}{255},...$and $S_1=frac{2}{3}, S_2=frac{14}{15} , S_3=frac{254}{255}, ...$
So I thought that this sum will converge to 1.
Let $$2^{2^{n-1}}=A_n, a_n=frac{A_n}{A_n^2-1}=frac{1}{2}left(frac{1}{A_n+1}+frac{1}{A_n-1}right)$$
Since $A_n = A_{n-1}^2$, I thought that this could lead to finding the partial sum of the series. I could prove that $$S_n=1-frac{1}{A_n^2-1}$$by mathematical induction, but I want to know whether my method can be useful for finding $S_n$ from scratch.
So my question is : Is there a way to find $S_n$ from scratch??
Also, I found that $$a_n=frac{frac{1}{A_n}}{1-frac{1}{A_n^2}}=sum_{n=1}^{infty}{frac{1}{A_n^{2n-1}}}=frac{1}{2^{1cdot2^{n-1}}}+frac{1}{2^{3cdot 2^{n-1}}}+frac{1}{2^{5cdot 2^{n-1}}}+....\
therefore sum_{n=1}^infty {a_n}=frac{1}{2^{1cdot 2^{0}}}+frac{1}{2^{3cdot 2^{0}}}+frac{1}{2^{5cdot 2^{0}}}+...\
+frac{1}{2^{1cdot 2^{1}}}+frac{1}{2^{3cdot 2^{1}}}+frac{1}{2^{5cdot 2^{1}}}+...\ +frac{1}{2^{1cdot 2^{2}}}+frac{1}{2^{3cdot 2^{2}}}+frac{1}{2^{5cdot 2^{2}}}+...
\ =sum_{n=1}^{infty}{2^{-n}}=1$$
Is this a valid way? I used the Riemann series theorem to rearrange them.
sequences-and-series
asked Jan 17 at 14:18
John. PJohn. P
1789
$endgroup$
A friend of mine gave me this quiz: Where does this sum converge to?
$$sumlimits_{n=1}^{infty}{frac{2^{2^{n-1}}}{2^{2^n}-1}}$$
$a_1=frac{2}{3} , a_2=frac{4}{15}, a_3=frac{16}{255},...$and $S_1=frac{2}{3}, S_2=frac{14}{15} , S_3=frac{254}{255}, ...$
So I thought that this sum will converge to 1.
Let $$2^{2^{n-1}}=A_n, a_n=frac{A_n}{A_n^2-1}=frac{1}{2}left(frac{1}{A_n+1}+frac{1}{A_n-1}right)$$
Since $A_n = A_{n-1}^2$, I thought that this could lead to finding the partial sum of the series. I could prove that $$S_n=1-frac{1}{A_n^2-1}$$by mathematical induction, but I want to know whether my method can be useful for finding $S_n$ from scratch.
So my question is : Is there a way to find $S_n$ from scratch??
Also, I found that $$a_n=frac{frac{1}{A_n}}{1-frac{1}{A_n^2}}=sum_{n=1}^{infty}{frac{1}{A_n^{2n-1}}}=frac{1}{2^{1cdot2^{n-1}}}+frac{1}{2^{3cdot 2^{n-1}}}+frac{1}{2^{5cdot 2^{n-1}}}+....\
therefore sum_{n=1}^infty {a_n}=frac{1}{2^{1cdot 2^{0}}}+frac{1}{2^{3cdot 2^{0}}}+frac{1}{2^{5cdot 2^{0}}}+...\
+frac{1}{2^{1cdot 2^{1}}}+frac{1}{2^{3cdot 2^{1}}}+frac{1}{2^{5cdot 2^{1}}}+...\ +frac{1}{2^{1cdot 2^{2}}}+frac{1}{2^{3cdot 2^{2}}}+frac{1}{2^{5cdot 2^{2}}}+...
\ =sum_{n=1}^{infty}{2^{-n}}=1$$
Is this a valid way? I used the Riemann series theorem to rearrange them.
sequences-and-series
sequences-and-series
asked Jan 17 at 14:18
John. PJohn. P
1789
asked Jan 17 at 14:18
John. PJohn. P
1789
edited Jan 17 at 23:37
John. P
asked Jan 17 at 14:18
John. PJohn. P
1789
asked Jan 17 at 14:18
John. PJohn. P
1789
asked Jan 17 at 14:18
John. PJohn. P
1789
1789
1
$begingroup$
This is valid, since the elements are non-negative,
$endgroup$
– Bonbon
Jan 17 at 14:23
$begingroup$
@Bonbon Thx for replying!!
$endgroup$
– John. P
Jan 17 at 15:24
$begingroup$
It is also interesting to show that $$ 1 = sum_{kgeq 0}frac{2^k}{2^{2^k}+1}.$$
$endgroup$
– Jack D'Aurizio
Jan 17 at 19:19
$begingroup$
Have a look at this and replace $z=frac{1}{2}$.
$endgroup$
– rtybase
Jan 17 at 21:41
$begingroup$
Notice that $frac{2^{2^{n-1}}}{2^{2^n}-1} = frac{2^{2^{n-1}}}{(2^{2^{n-1}}-1)(2^{2^{n-1}}+1)} = frac{1}{2^{2^{n-1}}-1}-frac{1}{2^{2^n}-1} equiv b_{n-1} - b_n$ where $b_n = frac{1}{2^{2^n}-1}$. Telescoping!
$endgroup$
– Winther
Jan 17 at 23:49
|
show 1 more comment
1
$begingroup$
This is valid, since the elements are non-negative,
$endgroup$
– Bonbon
Jan 17 at 14:23
$begingroup$
@Bonbon Thx for replying!!
$endgroup$
– John. P
Jan 17 at 15:24
$begingroup$
It is also interesting to show that $$ 1 = sum_{kgeq 0}frac{2^k}{2^{2^k}+1}.$$
$endgroup$
– Jack D'Aurizio
Jan 17 at 19:19
$begingroup$
Have a look at this and replace $z=frac{1}{2}$.
$endgroup$
– rtybase
Jan 17 at 21:41
$begingroup$
Notice that $frac{2^{2^{n-1}}}{2^{2^n}-1} = frac{2^{2^{n-1}}}{(2^{2^{n-1}}-1)(2^{2^{n-1}}+1)} = frac{1}{2^{2^{n-1}}-1}-frac{1}{2^{2^n}-1} equiv b_{n-1} - b_n$ where $b_n = frac{1}{2^{2^n}-1}$. Telescoping!
$endgroup$
– Winther
Jan 17 at 23:49
1
1
$begingroup$
This is valid, since the elements are non-negative,
$endgroup$
– Bonbon
Jan 17 at 14:23
$begingroup$
This is valid, since the elements are non-negative,
$endgroup$
– Bonbon
Jan 17 at 14:23
$begingroup$
@Bonbon Thx for replying!!
$endgroup$
– John. P
Jan 17 at 15:24
$begingroup$
@Bonbon Thx for replying!!
$endgroup$
– John. P
Jan 17 at 15:24
$begingroup$
It is also interesting to show that $$ 1 = sum_{kgeq 0}frac{2^k}{2^{2^k}+1}.$$
$endgroup$
– Jack D'Aurizio
Jan 17 at 19:19
$begingroup$
It is also interesting to show that $$ 1 = sum_{kgeq 0}frac{2^k}{2^{2^k}+1}.$$
$endgroup$
– Jack D'Aurizio
Jan 17 at 19:19
$begingroup$
Have a look at this and replace $z=frac{1}{2}$.
$endgroup$
– rtybase
Jan 17 at 21:41
$begingroup$
Have a look at this and replace $z=frac{1}{2}$.
$endgroup$
– rtybase
Jan 17 at 21:41
$begingroup$
Notice that $frac{2^{2^{n-1}}}{2^{2^n}-1} = frac{2^{2^{n-1}}}{(2^{2^{n-1}}-1)(2^{2^{n-1}}+1)} = frac{1}{2^{2^{n-1}}-1}-frac{1}{2^{2^n}-1} equiv b_{n-1} - b_n$ where $b_n = frac{1}{2^{2^n}-1}$. Telescoping!
$endgroup$
– Winther
Jan 17 at 23:49
$begingroup$
Notice that $frac{2^{2^{n-1}}}{2^{2^n}-1} = frac{2^{2^{n-1}}}{(2^{2^{n-1}}-1)(2^{2^{n-1}}+1)} = frac{1}{2^{2^{n-1}}-1}-frac{1}{2^{2^n}-1} equiv b_{n-1} - b_n$ where $b_n = frac{1}{2^{2^n}-1}$. Telescoping!
$endgroup$
– Winther
Jan 17 at 23:49
|
show 1 more comment
1 Answer
1
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oldest
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$begingroup$
Here's another manipulation you could do that doesn't require any rearranging, just the simple fact that two convergent series can be added term-wise (which is a commutativity of addition plus an addition of convergent sequences statement). Let's write $$s_n = frac{2^{2^{n-1}}}{2^{2^n}-1};$$
$$a_n = frac{1}{2^{2^{n-1}}-1};$$
$$b_n = frac{1}{2^{2^{n-1}}+1};$$
$S=sum_{n=1}^infty s_n$; $A = sum_{n=1}^infty a_n$; and $B = sum_{n=1}^infty b_n$. It's easy to see that $A,B < infty$ by comparison to some simple (e.g., geometric) series. Since $s_n = a_n/2 + b_n/2$, we have that $S=A/2+B/2<infty$ and the following hold:
$$a_n = a_{n-1}/2 - b_{n-1}/2, ;;;mbox{so}$$
$$A = a_{0}/2-b_0/2 + A/2 - B/2 = 1+A/2-B/2, ;;;mbox{so}$$
$$S = A/2 + B/2 = (1+A/2-B/2)/2+B/2 = 1/2+(A/2+B/2)/2 = 1/2 + S/2,$$
which of course implies $S=1$.
answered Jan 17 at 15:49
cspruncsprun
1,54828
$endgroup$
2
$begingroup$
How did you come up with this idea?? So brilliant!
$endgroup$
– John. P
Jan 17 at 15:57
add a comment |
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$begingroup$
Here's another manipulation you could do that doesn't require any rearranging, just the simple fact that two convergent series can be added term-wise (which is a commutativity of addition plus an addition of convergent sequences statement). Let's write $$s_n = frac{2^{2^{n-1}}}{2^{2^n}-1};$$
$$a_n = frac{1}{2^{2^{n-1}}-1};$$
$$b_n = frac{1}{2^{2^{n-1}}+1};$$
$S=sum_{n=1}^infty s_n$; $A = sum_{n=1}^infty a_n$; and $B = sum_{n=1}^infty b_n$. It's easy to see that $A,B < infty$ by comparison to some simple (e.g., geometric) series. Since $s_n = a_n/2 + b_n/2$, we have that $S=A/2+B/2<infty$ and the following hold:
$$a_n = a_{n-1}/2 - b_{n-1}/2, ;;;mbox{so}$$
$$A = a_{0}/2-b_0/2 + A/2 - B/2 = 1+A/2-B/2, ;;;mbox{so}$$
$$S = A/2 + B/2 = (1+A/2-B/2)/2+B/2 = 1/2+(A/2+B/2)/2 = 1/2 + S/2,$$
which of course implies $S=1$.
answered Jan 17 at 15:49
cspruncsprun
1,54828
$endgroup$
2
$begingroup$
How did you come up with this idea?? So brilliant!
$endgroup$
– John. P
Jan 17 at 15:57
add a comment |
$begingroup$
Here's another manipulation you could do that doesn't require any rearranging, just the simple fact that two convergent series can be added term-wise (which is a commutativity of addition plus an addition of convergent sequences statement). Let's write $$s_n = frac{2^{2^{n-1}}}{2^{2^n}-1};$$
$$a_n = frac{1}{2^{2^{n-1}}-1};$$
$$b_n = frac{1}{2^{2^{n-1}}+1};$$
$S=sum_{n=1}^infty s_n$; $A = sum_{n=1}^infty a_n$; and $B = sum_{n=1}^infty b_n$. It's easy to see that $A,B < infty$ by comparison to some simple (e.g., geometric) series. Since $s_n = a_n/2 + b_n/2$, we have that $S=A/2+B/2<infty$ and the following hold:
$$a_n = a_{n-1}/2 - b_{n-1}/2, ;;;mbox{so}$$
$$A = a_{0}/2-b_0/2 + A/2 - B/2 = 1+A/2-B/2, ;;;mbox{so}$$
$$S = A/2 + B/2 = (1+A/2-B/2)/2+B/2 = 1/2+(A/2+B/2)/2 = 1/2 + S/2,$$
which of course implies $S=1$.
answered Jan 17 at 15:49
cspruncsprun
1,54828
$endgroup$
2
$begingroup$
How did you come up with this idea?? So brilliant!
$endgroup$
– John. P
Jan 17 at 15:57
add a comment |
$begingroup$
Here's another manipulation you could do that doesn't require any rearranging, just the simple fact that two convergent series can be added term-wise (which is a commutativity of addition plus an addition of convergent sequences statement). Let's write $$s_n = frac{2^{2^{n-1}}}{2^{2^n}-1};$$
$$a_n = frac{1}{2^{2^{n-1}}-1};$$
$$b_n = frac{1}{2^{2^{n-1}}+1};$$
$S=sum_{n=1}^infty s_n$; $A = sum_{n=1}^infty a_n$; and $B = sum_{n=1}^infty b_n$. It's easy to see that $A,B < infty$ by comparison to some simple (e.g., geometric) series. Since $s_n = a_n/2 + b_n/2$, we have that $S=A/2+B/2<infty$ and the following hold:
$$a_n = a_{n-1}/2 - b_{n-1}/2, ;;;mbox{so}$$
$$A = a_{0}/2-b_0/2 + A/2 - B/2 = 1+A/2-B/2, ;;;mbox{so}$$
$$S = A/2 + B/2 = (1+A/2-B/2)/2+B/2 = 1/2+(A/2+B/2)/2 = 1/2 + S/2,$$
which of course implies $S=1$.
answered Jan 17 at 15:49
cspruncsprun
1,54828
$endgroup$
Here's another manipulation you could do that doesn't require any rearranging, just the simple fact that two convergent series can be added term-wise (which is a commutativity of addition plus an addition of convergent sequences statement). Let's write $$s_n = frac{2^{2^{n-1}}}{2^{2^n}-1};$$
$$a_n = frac{1}{2^{2^{n-1}}-1};$$
$$b_n = frac{1}{2^{2^{n-1}}+1};$$
$S=sum_{n=1}^infty s_n$; $A = sum_{n=1}^infty a_n$; and $B = sum_{n=1}^infty b_n$. It's easy to see that $A,B < infty$ by comparison to some simple (e.g., geometric) series. Since $s_n = a_n/2 + b_n/2$, we have that $S=A/2+B/2<infty$ and the following hold:
$$a_n = a_{n-1}/2 - b_{n-1}/2, ;;;mbox{so}$$
$$A = a_{0}/2-b_0/2 + A/2 - B/2 = 1+A/2-B/2, ;;;mbox{so}$$
$$S = A/2 + B/2 = (1+A/2-B/2)/2+B/2 = 1/2+(A/2+B/2)/2 = 1/2 + S/2,$$
which of course implies $S=1$.
answered Jan 17 at 15:49
cspruncsprun
1,54828
edited Jan 17 at 15:58
answered Jan 17 at 15:49
cspruncsprun
1,54828
answered Jan 17 at 15:49
cspruncsprun
1,54828
answered Jan 17 at 15:49
cspruncsprun
1,54828
1,54828
2
$begingroup$
How did you come up with this idea?? So brilliant!
$endgroup$
– John. P
Jan 17 at 15:57
add a comment |
2
$begingroup$
How did you come up with this idea?? So brilliant!
$endgroup$
– John. P
Jan 17 at 15:57
2
2
$begingroup$
How did you come up with this idea?? So brilliant!
$endgroup$
– John. P
Jan 17 at 15:57
$begingroup$
How did you come up with this idea?? So brilliant!
$endgroup$
– John. P
Jan 17 at 15:57
add a comment |
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$begingroup$
This is valid, since the elements are non-negative,
$endgroup$
– Bonbon
Jan 17 at 14:23
$begingroup$
@Bonbon Thx for replying!!
$endgroup$
– John. P
Jan 17 at 15:24
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It is also interesting to show that $$ 1 = sum_{kgeq 0}frac{2^k}{2^{2^k}+1}.$$
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– Jack D'Aurizio
Jan 17 at 19:19
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Have a look at this and replace $z=frac{1}{2}$.
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– rtybase
Jan 17 at 21:41
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Notice that $frac{2^{2^{n-1}}}{2^{2^n}-1} = frac{2^{2^{n-1}}}{(2^{2^{n-1}}-1)(2^{2^{n-1}}+1)} = frac{1}{2^{2^{n-1}}-1}-frac{1}{2^{2^n}-1} equiv b_{n-1} - b_n$ where $b_n = frac{1}{2^{2^n}-1}$. Telescoping!
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– Winther
Jan 17 at 23:49