Mixing
Alternating versions of Convergent Series summing to half
$begingroup$
Using $zeta(2)$ as an example.
$sum_{n=0}^infty frac{1}{n^2}=1+frac{1}{2^2}+frac{1}{3^2}+...=frac{pi^2}{6} $
When alternating the values gave -$frac{zeta(2)}{2}$
$sum_{n=0}^infty frac{(-1)^n}{n^2}=-1+frac{1}{2^2}-frac{1}{3^2}+...=-frac{pi^2}{12} $
and changing the negative gave $frac{zeta(2)}{2}$
$sum_{n=0}^infty frac{(-1)^n}{n^2}=1-frac{1}{2^2}+frac{1}{3^2}-...=frac{pi^2}{12} $
This seems to only work on convergent series, since something like the Harmonic Series continues to infinity while the Alternating Harmonic Series converges to ln(2)
So why do the alternating versions produce these values so similar to the non alternating version?
sequences-and-series convergence
asked Jan 30 at 0:44
2AM Aesthetic2AM Aesthetic
83
$endgroup$
add a comment |
$begingroup$
Using $zeta(2)$ as an example.
$sum_{n=0}^infty frac{1}{n^2}=1+frac{1}{2^2}+frac{1}{3^2}+...=frac{pi^2}{6} $
When alternating the values gave -$frac{zeta(2)}{2}$
$sum_{n=0}^infty frac{(-1)^n}{n^2}=-1+frac{1}{2^2}-frac{1}{3^2}+...=-frac{pi^2}{12} $
and changing the negative gave $frac{zeta(2)}{2}$
$sum_{n=0}^infty frac{(-1)^n}{n^2}=1-frac{1}{2^2}+frac{1}{3^2}-...=frac{pi^2}{12} $
This seems to only work on convergent series, since something like the Harmonic Series continues to infinity while the Alternating Harmonic Series converges to ln(2)
So why do the alternating versions produce these values so similar to the non alternating version?
sequences-and-series convergence
asked Jan 30 at 0:44
2AM Aesthetic2AM Aesthetic
83
$endgroup$
1
$begingroup$
Because the series using only even $n$ is exactly $(1/4)$ times the original series. Even: $pi^2/24.$ Therefore Odd: what is left over from $pi^2/6,$ namely $pi^2/8$
$endgroup$
– Will Jagy
Jan 30 at 0:49
add a comment |
$begingroup$
Using $zeta(2)$ as an example.
$sum_{n=0}^infty frac{1}{n^2}=1+frac{1}{2^2}+frac{1}{3^2}+...=frac{pi^2}{6} $
When alternating the values gave -$frac{zeta(2)}{2}$
$sum_{n=0}^infty frac{(-1)^n}{n^2}=-1+frac{1}{2^2}-frac{1}{3^2}+...=-frac{pi^2}{12} $
and changing the negative gave $frac{zeta(2)}{2}$
$sum_{n=0}^infty frac{(-1)^n}{n^2}=1-frac{1}{2^2}+frac{1}{3^2}-...=frac{pi^2}{12} $
This seems to only work on convergent series, since something like the Harmonic Series continues to infinity while the Alternating Harmonic Series converges to ln(2)
So why do the alternating versions produce these values so similar to the non alternating version?
sequences-and-series convergence
asked Jan 30 at 0:44
2AM Aesthetic2AM Aesthetic
83
$endgroup$
Using $zeta(2)$ as an example.
$sum_{n=0}^infty frac{1}{n^2}=1+frac{1}{2^2}+frac{1}{3^2}+...=frac{pi^2}{6} $
When alternating the values gave -$frac{zeta(2)}{2}$
$sum_{n=0}^infty frac{(-1)^n}{n^2}=-1+frac{1}{2^2}-frac{1}{3^2}+...=-frac{pi^2}{12} $
and changing the negative gave $frac{zeta(2)}{2}$
$sum_{n=0}^infty frac{(-1)^n}{n^2}=1-frac{1}{2^2}+frac{1}{3^2}-...=frac{pi^2}{12} $
This seems to only work on convergent series, since something like the Harmonic Series continues to infinity while the Alternating Harmonic Series converges to ln(2)
So why do the alternating versions produce these values so similar to the non alternating version?
sequences-and-series convergence
sequences-and-series convergence
asked Jan 30 at 0:44
2AM Aesthetic2AM Aesthetic
83
asked Jan 30 at 0:44
2AM Aesthetic2AM Aesthetic
83
asked Jan 30 at 0:44
2AM Aesthetic2AM Aesthetic
83
asked Jan 30 at 0:44
2AM Aesthetic2AM Aesthetic
83
asked Jan 30 at 0:44
2AM Aesthetic2AM Aesthetic
83
83
1
$begingroup$
Because the series using only even $n$ is exactly $(1/4)$ times the original series. Even: $pi^2/24.$ Therefore Odd: what is left over from $pi^2/6,$ namely $pi^2/8$
$endgroup$
– Will Jagy
Jan 30 at 0:49
add a comment |
1
$begingroup$
Because the series using only even $n$ is exactly $(1/4)$ times the original series. Even: $pi^2/24.$ Therefore Odd: what is left over from $pi^2/6,$ namely $pi^2/8$
$endgroup$
– Will Jagy
Jan 30 at 0:49
1
1
$begingroup$
Because the series using only even $n$ is exactly $(1/4)$ times the original series. Even: $pi^2/24.$ Therefore Odd: what is left over from $pi^2/6,$ namely $pi^2/8$
$endgroup$
– Will Jagy
Jan 30 at 0:49
$begingroup$
Because the series using only even $n$ is exactly $(1/4)$ times the original series. Even: $pi^2/24.$ Therefore Odd: what is left over from $pi^2/6,$ namely $pi^2/8$
$endgroup$
– Will Jagy
Jan 30 at 0:49
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that we have in general
$$begin{align}
sum_{n=1}^{2N} (-1)^na_n&=sum_{n=1}^N a_{2n}-sum_{n=1}^Na_{2n-1}\\
&=sum_{n=1}^N a_{2n}-left(sum_{n=1}^{2N}a_n-sum_{n=1}^Na_{2n}right)\\
&=2sum_{n=1}^N a_{2n}-sum_{n=1}^{2N} a_ntag1
end{align}$$
We can use $(1)$ to evaluate the series of interest. Using $(1)$, for $a_n=frac1{n^2}$ we have
$$begin{align}sum_{n=1}^{2N}frac{(-1)^n}{n^2}&=2sum_{n=1}^N frac1{(2n)^2}-sum_{n=1}^{2N}frac1{n^2}\\
&=-frac12sum_{n=1}^N frac1{n^2}-sum_{n=N+1}^{2N}frac1{n^2}tag2
end{align}$$
Since the first term on the right-hand side of $(2)$ converges to $-frac{pi^2}{12}$ while the second term converges to $0$, we obtain the expected result
$$sum_{n=1}^infty frac{(-1)^n}{n^2}=-frac{pi^2}{12}$$
Note that the expression in $(1)$ can be useful in evaluating other series.
EXAMPLE $1$:
For example, in THIS ANSWER, I used $(1)$ with $a_n=frac{log(n)}{n}$ to show that
$$sum_{n=1}^infty frac{(-1)^nlog(n)}{n}=gammalog(2)-frac12log^2(2)$$
EXAMPLE $2$:
As another example of $(1)$, let $a_n=frac1n$. Then, we see that
$$begin{align}
sum_{n=1}^{2N}frac{(-1)^n}{n}&=2sum_{n=1}^N frac1{2n}-sum_{n=1}^{2N}frac1{n}\\
&=-sum_{n=N+1}^{2N}frac1n\\
&=-frac1Nsum_{n=1}^N frac1{1+(n/N)}tag3
end{align}$$
The right-hand side of $(3)$ is the Riemann sum for $-int_0^1frac1{1+x},dx=-log(2)$. Therefore, we find that
$$sum_{n=1}^infty frac{(-1)^n}{n}=-log(2)$$
answered Jan 30 at 3:33
Mark ViolaMark Viola
134k1278176
$endgroup$
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jan 30 at 5:02
add a comment |
$begingroup$
The alternating series you present is $eta(2)$, the Dirichlet eta function. More specifically,
$$eta(s) = sum_{n=1}^infty frac{(-1)^{n-1}}{n^s}.$$
It turns out that this series, unlike the zeta series, converges for all $s gt 0$. We can produce a very special relationship to the zeta function $zeta(s)$ with some algebraic cleverness:
begin{align} sum_{n=1}^infty frac{(-1)^{n-1}}{n^s} &= 1 - 2^{-s} + 3^{-s} - 4^{-s} + 5^{-s} - 6^{-s} + 7^{-s} - 8^{-s} + cdots \
&= left( 1 + 3^{-s} + 5^{-s} + 7^{-s} + cdots right) - left( 2^{-s} + 4^{-s} + 6^{-s} + 8^{-s} + cdots right) \
&= left( 1 + 2^{-s} + 3^{-s} + 4^{-s} + 5^{-s} + 6^{-s} + 7^{-s} + cdots right) - 2 left( 2^{-s} + 4^{-s} + 6^{-s} + 8^{-s} + cdots right) \
&= left( 1 + 2^{-s} + 3^{-s} + 4^{-s} + 5^{-s} + 6^{-s} + 7^{-s} + cdots right) - 2 left( 2^{-s} 1 + 2^{-s} 2^{-s} + 2^{-s} 3^{-s} + 2^{-s} 4^{-s} + cdots right) \
&= zeta(s) - 2^{1 - s} left(1 + 2^{-s} + 3^{-s} + 4^{-s} + cdots right) \
&= zeta(s) - 2^{1 - s} zeta(s) \
&= left( 1 - 2^{1 - s} right) zeta(s).
end{align}
So now, by our new formula $eta(s) = left( 1 - 2^{1 - s} right) zeta(s)$, we can produce the value of the eta function given the corresponding zeta function, which will differ only by a rational factor.
So if $zeta(2) = frac{pi^2}6$, then $eta(2) = left( 1 - 2^{-1} right) zeta(2) = frac12 frac{pi^2}6 = frac{pi^2}{12},$ as desired.
This formula can also be manipulated into the form $zeta(s) = frac{eta(s)}{1 - 2^{1 - s}}$, which allows us to find a value for zeta in $0 lt s lt 1$.
The reason the series for $eta(1)$ converges (conditionally) is at $s = 1$, that the factor $1 - 2^{1 - s}$ has a simple (i.e. degree one) zero, while $zeta(s)$ has a simple pole. These two evenly matched factors are battling for control, and their fight lands them at this crossroad. (It's a bit more complicated to show that it converges to $ln 2$, but it can be done.)
answered Jan 30 at 4:30
BladewoodBladewood
335213
$endgroup$
add a comment |
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2 Answers
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$begingroup$
Note that we have in general
$$begin{align}
sum_{n=1}^{2N} (-1)^na_n&=sum_{n=1}^N a_{2n}-sum_{n=1}^Na_{2n-1}\\
&=sum_{n=1}^N a_{2n}-left(sum_{n=1}^{2N}a_n-sum_{n=1}^Na_{2n}right)\\
&=2sum_{n=1}^N a_{2n}-sum_{n=1}^{2N} a_ntag1
end{align}$$
We can use $(1)$ to evaluate the series of interest. Using $(1)$, for $a_n=frac1{n^2}$ we have
$$begin{align}sum_{n=1}^{2N}frac{(-1)^n}{n^2}&=2sum_{n=1}^N frac1{(2n)^2}-sum_{n=1}^{2N}frac1{n^2}\\
&=-frac12sum_{n=1}^N frac1{n^2}-sum_{n=N+1}^{2N}frac1{n^2}tag2
end{align}$$
Since the first term on the right-hand side of $(2)$ converges to $-frac{pi^2}{12}$ while the second term converges to $0$, we obtain the expected result
$$sum_{n=1}^infty frac{(-1)^n}{n^2}=-frac{pi^2}{12}$$
Note that the expression in $(1)$ can be useful in evaluating other series.
EXAMPLE $1$:
For example, in THIS ANSWER, I used $(1)$ with $a_n=frac{log(n)}{n}$ to show that
$$sum_{n=1}^infty frac{(-1)^nlog(n)}{n}=gammalog(2)-frac12log^2(2)$$
EXAMPLE $2$:
As another example of $(1)$, let $a_n=frac1n$. Then, we see that
$$begin{align}
sum_{n=1}^{2N}frac{(-1)^n}{n}&=2sum_{n=1}^N frac1{2n}-sum_{n=1}^{2N}frac1{n}\\
&=-sum_{n=N+1}^{2N}frac1n\\
&=-frac1Nsum_{n=1}^N frac1{1+(n/N)}tag3
end{align}$$
The right-hand side of $(3)$ is the Riemann sum for $-int_0^1frac1{1+x},dx=-log(2)$. Therefore, we find that
$$sum_{n=1}^infty frac{(-1)^n}{n}=-log(2)$$
answered Jan 30 at 3:33
Mark ViolaMark Viola
134k1278176
$endgroup$
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jan 30 at 5:02
add a comment |
$begingroup$
Note that we have in general
$$begin{align}
sum_{n=1}^{2N} (-1)^na_n&=sum_{n=1}^N a_{2n}-sum_{n=1}^Na_{2n-1}\\
&=sum_{n=1}^N a_{2n}-left(sum_{n=1}^{2N}a_n-sum_{n=1}^Na_{2n}right)\\
&=2sum_{n=1}^N a_{2n}-sum_{n=1}^{2N} a_ntag1
end{align}$$
We can use $(1)$ to evaluate the series of interest. Using $(1)$, for $a_n=frac1{n^2}$ we have
$$begin{align}sum_{n=1}^{2N}frac{(-1)^n}{n^2}&=2sum_{n=1}^N frac1{(2n)^2}-sum_{n=1}^{2N}frac1{n^2}\\
&=-frac12sum_{n=1}^N frac1{n^2}-sum_{n=N+1}^{2N}frac1{n^2}tag2
end{align}$$
Since the first term on the right-hand side of $(2)$ converges to $-frac{pi^2}{12}$ while the second term converges to $0$, we obtain the expected result
$$sum_{n=1}^infty frac{(-1)^n}{n^2}=-frac{pi^2}{12}$$
Note that the expression in $(1)$ can be useful in evaluating other series.
EXAMPLE $1$:
For example, in THIS ANSWER, I used $(1)$ with $a_n=frac{log(n)}{n}$ to show that
$$sum_{n=1}^infty frac{(-1)^nlog(n)}{n}=gammalog(2)-frac12log^2(2)$$
EXAMPLE $2$:
As another example of $(1)$, let $a_n=frac1n$. Then, we see that
$$begin{align}
sum_{n=1}^{2N}frac{(-1)^n}{n}&=2sum_{n=1}^N frac1{2n}-sum_{n=1}^{2N}frac1{n}\\
&=-sum_{n=N+1}^{2N}frac1n\\
&=-frac1Nsum_{n=1}^N frac1{1+(n/N)}tag3
end{align}$$
The right-hand side of $(3)$ is the Riemann sum for $-int_0^1frac1{1+x},dx=-log(2)$. Therefore, we find that
$$sum_{n=1}^infty frac{(-1)^n}{n}=-log(2)$$
answered Jan 30 at 3:33
Mark ViolaMark Viola
134k1278176
$endgroup$
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jan 30 at 5:02
add a comment |
$begingroup$
Note that we have in general
$$begin{align}
sum_{n=1}^{2N} (-1)^na_n&=sum_{n=1}^N a_{2n}-sum_{n=1}^Na_{2n-1}\\
&=sum_{n=1}^N a_{2n}-left(sum_{n=1}^{2N}a_n-sum_{n=1}^Na_{2n}right)\\
&=2sum_{n=1}^N a_{2n}-sum_{n=1}^{2N} a_ntag1
end{align}$$
We can use $(1)$ to evaluate the series of interest. Using $(1)$, for $a_n=frac1{n^2}$ we have
$$begin{align}sum_{n=1}^{2N}frac{(-1)^n}{n^2}&=2sum_{n=1}^N frac1{(2n)^2}-sum_{n=1}^{2N}frac1{n^2}\\
&=-frac12sum_{n=1}^N frac1{n^2}-sum_{n=N+1}^{2N}frac1{n^2}tag2
end{align}$$
Since the first term on the right-hand side of $(2)$ converges to $-frac{pi^2}{12}$ while the second term converges to $0$, we obtain the expected result
$$sum_{n=1}^infty frac{(-1)^n}{n^2}=-frac{pi^2}{12}$$
Note that the expression in $(1)$ can be useful in evaluating other series.
EXAMPLE $1$:
For example, in THIS ANSWER, I used $(1)$ with $a_n=frac{log(n)}{n}$ to show that
$$sum_{n=1}^infty frac{(-1)^nlog(n)}{n}=gammalog(2)-frac12log^2(2)$$
EXAMPLE $2$:
As another example of $(1)$, let $a_n=frac1n$. Then, we see that
$$begin{align}
sum_{n=1}^{2N}frac{(-1)^n}{n}&=2sum_{n=1}^N frac1{2n}-sum_{n=1}^{2N}frac1{n}\\
&=-sum_{n=N+1}^{2N}frac1n\\
&=-frac1Nsum_{n=1}^N frac1{1+(n/N)}tag3
end{align}$$
The right-hand side of $(3)$ is the Riemann sum for $-int_0^1frac1{1+x},dx=-log(2)$. Therefore, we find that
$$sum_{n=1}^infty frac{(-1)^n}{n}=-log(2)$$
answered Jan 30 at 3:33
Mark ViolaMark Viola
134k1278176
$endgroup$
Note that we have in general
$$begin{align}
sum_{n=1}^{2N} (-1)^na_n&=sum_{n=1}^N a_{2n}-sum_{n=1}^Na_{2n-1}\\
&=sum_{n=1}^N a_{2n}-left(sum_{n=1}^{2N}a_n-sum_{n=1}^Na_{2n}right)\\
&=2sum_{n=1}^N a_{2n}-sum_{n=1}^{2N} a_ntag1
end{align}$$
We can use $(1)$ to evaluate the series of interest. Using $(1)$, for $a_n=frac1{n^2}$ we have
$$begin{align}sum_{n=1}^{2N}frac{(-1)^n}{n^2}&=2sum_{n=1}^N frac1{(2n)^2}-sum_{n=1}^{2N}frac1{n^2}\\
&=-frac12sum_{n=1}^N frac1{n^2}-sum_{n=N+1}^{2N}frac1{n^2}tag2
end{align}$$
Since the first term on the right-hand side of $(2)$ converges to $-frac{pi^2}{12}$ while the second term converges to $0$, we obtain the expected result
$$sum_{n=1}^infty frac{(-1)^n}{n^2}=-frac{pi^2}{12}$$
Note that the expression in $(1)$ can be useful in evaluating other series.
EXAMPLE $1$:
For example, in THIS ANSWER, I used $(1)$ with $a_n=frac{log(n)}{n}$ to show that
$$sum_{n=1}^infty frac{(-1)^nlog(n)}{n}=gammalog(2)-frac12log^2(2)$$
EXAMPLE $2$:
As another example of $(1)$, let $a_n=frac1n$. Then, we see that
$$begin{align}
sum_{n=1}^{2N}frac{(-1)^n}{n}&=2sum_{n=1}^N frac1{2n}-sum_{n=1}^{2N}frac1{n}\\
&=-sum_{n=N+1}^{2N}frac1n\\
&=-frac1Nsum_{n=1}^N frac1{1+(n/N)}tag3
end{align}$$
The right-hand side of $(3)$ is the Riemann sum for $-int_0^1frac1{1+x},dx=-log(2)$. Therefore, we find that
$$sum_{n=1}^infty frac{(-1)^n}{n}=-log(2)$$
answered Jan 30 at 3:33
Mark ViolaMark Viola
134k1278176
edited Jan 30 at 3:53
answered Jan 30 at 3:33
Mark ViolaMark Viola
134k1278176
answered Jan 30 at 3:33
Mark ViolaMark Viola
134k1278176
answered Jan 30 at 3:33
Mark ViolaMark Viola
134k1278176
134k1278176
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jan 30 at 5:02
add a comment |
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jan 30 at 5:02
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jan 30 at 5:02
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jan 30 at 5:02
add a comment |
$begingroup$
The alternating series you present is $eta(2)$, the Dirichlet eta function. More specifically,
$$eta(s) = sum_{n=1}^infty frac{(-1)^{n-1}}{n^s}.$$
It turns out that this series, unlike the zeta series, converges for all $s gt 0$. We can produce a very special relationship to the zeta function $zeta(s)$ with some algebraic cleverness:
begin{align} sum_{n=1}^infty frac{(-1)^{n-1}}{n^s} &= 1 - 2^{-s} + 3^{-s} - 4^{-s} + 5^{-s} - 6^{-s} + 7^{-s} - 8^{-s} + cdots \
&= left( 1 + 3^{-s} + 5^{-s} + 7^{-s} + cdots right) - left( 2^{-s} + 4^{-s} + 6^{-s} + 8^{-s} + cdots right) \
&= left( 1 + 2^{-s} + 3^{-s} + 4^{-s} + 5^{-s} + 6^{-s} + 7^{-s} + cdots right) - 2 left( 2^{-s} + 4^{-s} + 6^{-s} + 8^{-s} + cdots right) \
&= left( 1 + 2^{-s} + 3^{-s} + 4^{-s} + 5^{-s} + 6^{-s} + 7^{-s} + cdots right) - 2 left( 2^{-s} 1 + 2^{-s} 2^{-s} + 2^{-s} 3^{-s} + 2^{-s} 4^{-s} + cdots right) \
&= zeta(s) - 2^{1 - s} left(1 + 2^{-s} + 3^{-s} + 4^{-s} + cdots right) \
&= zeta(s) - 2^{1 - s} zeta(s) \
&= left( 1 - 2^{1 - s} right) zeta(s).
end{align}
So now, by our new formula $eta(s) = left( 1 - 2^{1 - s} right) zeta(s)$, we can produce the value of the eta function given the corresponding zeta function, which will differ only by a rational factor.
So if $zeta(2) = frac{pi^2}6$, then $eta(2) = left( 1 - 2^{-1} right) zeta(2) = frac12 frac{pi^2}6 = frac{pi^2}{12},$ as desired.
This formula can also be manipulated into the form $zeta(s) = frac{eta(s)}{1 - 2^{1 - s}}$, which allows us to find a value for zeta in $0 lt s lt 1$.
The reason the series for $eta(1)$ converges (conditionally) is at $s = 1$, that the factor $1 - 2^{1 - s}$ has a simple (i.e. degree one) zero, while $zeta(s)$ has a simple pole. These two evenly matched factors are battling for control, and their fight lands them at this crossroad. (It's a bit more complicated to show that it converges to $ln 2$, but it can be done.)
answered Jan 30 at 4:30
BladewoodBladewood
335213
$endgroup$
add a comment |
$begingroup$
The alternating series you present is $eta(2)$, the Dirichlet eta function. More specifically,
$$eta(s) = sum_{n=1}^infty frac{(-1)^{n-1}}{n^s}.$$
It turns out that this series, unlike the zeta series, converges for all $s gt 0$. We can produce a very special relationship to the zeta function $zeta(s)$ with some algebraic cleverness:
begin{align} sum_{n=1}^infty frac{(-1)^{n-1}}{n^s} &= 1 - 2^{-s} + 3^{-s} - 4^{-s} + 5^{-s} - 6^{-s} + 7^{-s} - 8^{-s} + cdots \
&= left( 1 + 3^{-s} + 5^{-s} + 7^{-s} + cdots right) - left( 2^{-s} + 4^{-s} + 6^{-s} + 8^{-s} + cdots right) \
&= left( 1 + 2^{-s} + 3^{-s} + 4^{-s} + 5^{-s} + 6^{-s} + 7^{-s} + cdots right) - 2 left( 2^{-s} + 4^{-s} + 6^{-s} + 8^{-s} + cdots right) \
&= left( 1 + 2^{-s} + 3^{-s} + 4^{-s} + 5^{-s} + 6^{-s} + 7^{-s} + cdots right) - 2 left( 2^{-s} 1 + 2^{-s} 2^{-s} + 2^{-s} 3^{-s} + 2^{-s} 4^{-s} + cdots right) \
&= zeta(s) - 2^{1 - s} left(1 + 2^{-s} + 3^{-s} + 4^{-s} + cdots right) \
&= zeta(s) - 2^{1 - s} zeta(s) \
&= left( 1 - 2^{1 - s} right) zeta(s).
end{align}
So now, by our new formula $eta(s) = left( 1 - 2^{1 - s} right) zeta(s)$, we can produce the value of the eta function given the corresponding zeta function, which will differ only by a rational factor.
So if $zeta(2) = frac{pi^2}6$, then $eta(2) = left( 1 - 2^{-1} right) zeta(2) = frac12 frac{pi^2}6 = frac{pi^2}{12},$ as desired.
This formula can also be manipulated into the form $zeta(s) = frac{eta(s)}{1 - 2^{1 - s}}$, which allows us to find a value for zeta in $0 lt s lt 1$.
The reason the series for $eta(1)$ converges (conditionally) is at $s = 1$, that the factor $1 - 2^{1 - s}$ has a simple (i.e. degree one) zero, while $zeta(s)$ has a simple pole. These two evenly matched factors are battling for control, and their fight lands them at this crossroad. (It's a bit more complicated to show that it converges to $ln 2$, but it can be done.)
answered Jan 30 at 4:30
BladewoodBladewood
335213
$endgroup$
add a comment |
$begingroup$
The alternating series you present is $eta(2)$, the Dirichlet eta function. More specifically,
$$eta(s) = sum_{n=1}^infty frac{(-1)^{n-1}}{n^s}.$$
It turns out that this series, unlike the zeta series, converges for all $s gt 0$. We can produce a very special relationship to the zeta function $zeta(s)$ with some algebraic cleverness:
begin{align} sum_{n=1}^infty frac{(-1)^{n-1}}{n^s} &= 1 - 2^{-s} + 3^{-s} - 4^{-s} + 5^{-s} - 6^{-s} + 7^{-s} - 8^{-s} + cdots \
&= left( 1 + 3^{-s} + 5^{-s} + 7^{-s} + cdots right) - left( 2^{-s} + 4^{-s} + 6^{-s} + 8^{-s} + cdots right) \
&= left( 1 + 2^{-s} + 3^{-s} + 4^{-s} + 5^{-s} + 6^{-s} + 7^{-s} + cdots right) - 2 left( 2^{-s} + 4^{-s} + 6^{-s} + 8^{-s} + cdots right) \
&= left( 1 + 2^{-s} + 3^{-s} + 4^{-s} + 5^{-s} + 6^{-s} + 7^{-s} + cdots right) - 2 left( 2^{-s} 1 + 2^{-s} 2^{-s} + 2^{-s} 3^{-s} + 2^{-s} 4^{-s} + cdots right) \
&= zeta(s) - 2^{1 - s} left(1 + 2^{-s} + 3^{-s} + 4^{-s} + cdots right) \
&= zeta(s) - 2^{1 - s} zeta(s) \
&= left( 1 - 2^{1 - s} right) zeta(s).
end{align}
So now, by our new formula $eta(s) = left( 1 - 2^{1 - s} right) zeta(s)$, we can produce the value of the eta function given the corresponding zeta function, which will differ only by a rational factor.
So if $zeta(2) = frac{pi^2}6$, then $eta(2) = left( 1 - 2^{-1} right) zeta(2) = frac12 frac{pi^2}6 = frac{pi^2}{12},$ as desired.
This formula can also be manipulated into the form $zeta(s) = frac{eta(s)}{1 - 2^{1 - s}}$, which allows us to find a value for zeta in $0 lt s lt 1$.
The reason the series for $eta(1)$ converges (conditionally) is at $s = 1$, that the factor $1 - 2^{1 - s}$ has a simple (i.e. degree one) zero, while $zeta(s)$ has a simple pole. These two evenly matched factors are battling for control, and their fight lands them at this crossroad. (It's a bit more complicated to show that it converges to $ln 2$, but it can be done.)
answered Jan 30 at 4:30
BladewoodBladewood
335213
$endgroup$
The alternating series you present is $eta(2)$, the Dirichlet eta function. More specifically,
$$eta(s) = sum_{n=1}^infty frac{(-1)^{n-1}}{n^s}.$$
It turns out that this series, unlike the zeta series, converges for all $s gt 0$. We can produce a very special relationship to the zeta function $zeta(s)$ with some algebraic cleverness:
begin{align} sum_{n=1}^infty frac{(-1)^{n-1}}{n^s} &= 1 - 2^{-s} + 3^{-s} - 4^{-s} + 5^{-s} - 6^{-s} + 7^{-s} - 8^{-s} + cdots \
&= left( 1 + 3^{-s} + 5^{-s} + 7^{-s} + cdots right) - left( 2^{-s} + 4^{-s} + 6^{-s} + 8^{-s} + cdots right) \
&= left( 1 + 2^{-s} + 3^{-s} + 4^{-s} + 5^{-s} + 6^{-s} + 7^{-s} + cdots right) - 2 left( 2^{-s} + 4^{-s} + 6^{-s} + 8^{-s} + cdots right) \
&= left( 1 + 2^{-s} + 3^{-s} + 4^{-s} + 5^{-s} + 6^{-s} + 7^{-s} + cdots right) - 2 left( 2^{-s} 1 + 2^{-s} 2^{-s} + 2^{-s} 3^{-s} + 2^{-s} 4^{-s} + cdots right) \
&= zeta(s) - 2^{1 - s} left(1 + 2^{-s} + 3^{-s} + 4^{-s} + cdots right) \
&= zeta(s) - 2^{1 - s} zeta(s) \
&= left( 1 - 2^{1 - s} right) zeta(s).
end{align}
So now, by our new formula $eta(s) = left( 1 - 2^{1 - s} right) zeta(s)$, we can produce the value of the eta function given the corresponding zeta function, which will differ only by a rational factor.
So if $zeta(2) = frac{pi^2}6$, then $eta(2) = left( 1 - 2^{-1} right) zeta(2) = frac12 frac{pi^2}6 = frac{pi^2}{12},$ as desired.
This formula can also be manipulated into the form $zeta(s) = frac{eta(s)}{1 - 2^{1 - s}}$, which allows us to find a value for zeta in $0 lt s lt 1$.
The reason the series for $eta(1)$ converges (conditionally) is at $s = 1$, that the factor $1 - 2^{1 - s}$ has a simple (i.e. degree one) zero, while $zeta(s)$ has a simple pole. These two evenly matched factors are battling for control, and their fight lands them at this crossroad. (It's a bit more complicated to show that it converges to $ln 2$, but it can be done.)
answered Jan 30 at 4:30
BladewoodBladewood
335213
answered Jan 30 at 4:30
BladewoodBladewood
335213
answered Jan 30 at 4:30
BladewoodBladewood
335213
answered Jan 30 at 4:30
BladewoodBladewood
335213
335213
add a comment |
add a comment |
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1
$begingroup$
Because the series using only even $n$ is exactly $(1/4)$ times the original series. Even: $pi^2/24.$ Therefore Odd: what is left over from $pi^2/6,$ namely $pi^2/8$
$endgroup$
– Will Jagy
Jan 30 at 0:49