Alternating series
- Tolaso J Kos
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Alternating series
Prove that:
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}=\frac{\pi^3}{32}$$
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}=\frac{\pi^3}{32}$$
Imagination is much more important than knowledge.
- Grigorios Kostakos
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Re: Alternating series
The sum \(\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}\) it is a special value of the Dirichlet's beta function
\[\beta(s) = \sum_{n=0}^\infty \frac{(-1)^n} {(2n+1)^s}\,.\]
It also holds: \[\beta(2k+1)={{{({-1})^k}{E_{2k}}{\pi^{2k+1}} \over {4^{k+1}}(2k)!}}\,,\quad k\in\mathbb{N}\,,\] where \(E_{2k}\) are the even-indexed Euler numbers.
For \(k=1\) we have that \[\beta(3)={{{({-1})^1}{E_{2}}{\pi^{3}} \over {4^{2}}2!}}=\frac{(-1)(-1)\pi^3}{32}=\frac{\pi^3}{32}\,.\] So \[\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}=\beta(3)=\frac{\pi^3}{32}\,.\]
P.S. Maybe there is a elementary proof for the equality using Fourier series. I'll be glad to see a such.
\[\beta(s) = \sum_{n=0}^\infty \frac{(-1)^n} {(2n+1)^s}\,.\]
It also holds: \[\beta(2k+1)={{{({-1})^k}{E_{2k}}{\pi^{2k+1}} \over {4^{k+1}}(2k)!}}\,,\quad k\in\mathbb{N}\,,\] where \(E_{2k}\) are the even-indexed Euler numbers.
For \(k=1\) we have that \[\beta(3)={{{({-1})^1}{E_{2}}{\pi^{3}} \over {4^{2}}2!}}=\frac{(-1)(-1)\pi^3}{32}=\frac{\pi^3}{32}\,.\] So \[\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}=\beta(3)=\frac{\pi^3}{32}\,.\]
P.S. Maybe there is a elementary proof for the equality using Fourier series. I'll be glad to see a such.
Grigorios Kostakos
- Tolaso J Kos
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Re: Alternating series
Yes, there exists...Grigorios Kostakos wrote: P.S. Maybe there is a elementary proof for the equality using Fourier series. I'll be glad to see a such.
Expanding the function \( f(x)= x(1-x), \; x \in [0, 1] \) in Fourier series and plugging \( x =1/2 \).
It is easy to note that \( f \) can be expressed in \( \sin \) Fourier series by:
$$f(x)=\frac{8}{\pi^3}\sum_{n=1}^{\infty} \frac{\sin (2n-1)\pi x }{(2n-1)^3}$$
Plugging \(x=1/2 \) we get what we want.
Imagination is much more important than knowledge.
- Tolaso J Kos
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Re: Alternating series
Indeed it holds that the sum is \( \beta(3) \) where \( \beta \) is the Beta Dirichlet.. Here is a somewhat neat solution:
$$\begin{aligned}
1-\frac{1}{3^3}+\frac{1}{5^3}-\cdots &=\sum_{n=0}^{\infty}\frac{(-1)^n}{\left ( 2n+1 \right )^3} \\
&\overset{(*)}{=} \left ( 1+\frac{1}{5^3}+\frac{1}{9^3}+\cdots \right )-\left ( \frac{1}{3^3}+\frac{1}{7^3}+\frac{1}{11^3}+\cdots \right )\\
&=\sum_{n=0}^{\infty}\frac{1}{\left ( 4n+1 \right )^3} \; -\sum_{n=0}^{\infty}\frac{1}{\left ( 4n+3 \right )^3} \\
&= -\frac{1}{2\cdot 4^3}\psi^{(2)}\left ( \frac{1}{4} \right )+\frac{1}{2\cdot 4^3}\psi^{(2)}\left ( \frac{3}{4} \right )\\
&=\frac{1}{2\cdot 4^3}\left [ \psi^{(2)}\left ( 1-\frac{1}{4} \right )-\psi^{(2)}\left ( \frac{1}{4} \right ) \right ]\\
&=\frac{1}{2\cdot 4^3}\left [ 2\pi^3 \cot \frac{\pi}{4} \csc^2 \frac{\pi}{4} \right ] \\
&=\frac{\pi^3 \cot \frac{\pi}{4}\csc^2 \frac{\pi}{4}}{4^3}=\frac{\pi^3}{32}
\end{aligned}$$
\( (*) \) due to absolute convergence.
$$\begin{aligned}
1-\frac{1}{3^3}+\frac{1}{5^3}-\cdots &=\sum_{n=0}^{\infty}\frac{(-1)^n}{\left ( 2n+1 \right )^3} \\
&\overset{(*)}{=} \left ( 1+\frac{1}{5^3}+\frac{1}{9^3}+\cdots \right )-\left ( \frac{1}{3^3}+\frac{1}{7^3}+\frac{1}{11^3}+\cdots \right )\\
&=\sum_{n=0}^{\infty}\frac{1}{\left ( 4n+1 \right )^3} \; -\sum_{n=0}^{\infty}\frac{1}{\left ( 4n+3 \right )^3} \\
&= -\frac{1}{2\cdot 4^3}\psi^{(2)}\left ( \frac{1}{4} \right )+\frac{1}{2\cdot 4^3}\psi^{(2)}\left ( \frac{3}{4} \right )\\
&=\frac{1}{2\cdot 4^3}\left [ \psi^{(2)}\left ( 1-\frac{1}{4} \right )-\psi^{(2)}\left ( \frac{1}{4} \right ) \right ]\\
&=\frac{1}{2\cdot 4^3}\left [ 2\pi^3 \cot \frac{\pi}{4} \csc^2 \frac{\pi}{4} \right ] \\
&=\frac{\pi^3 \cot \frac{\pi}{4}\csc^2 \frac{\pi}{4}}{4^3}=\frac{\pi^3}{32}
\end{aligned}$$
\( (*) \) due to absolute convergence.
Imagination is much more important than knowledge.
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