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 Post subject: An infinite sum
PostPosted: Sat Mar 11, 2017 8:40 pm 

Joined: Sat Nov 14, 2015 6:32 am
Posts: 126
Location: Melbourne, Australia
Let $r \in \mathbb{Z}$. Prove that

$$\sum_{n=-\infty}^{\infty} \arctan \left( \frac{\sinh r}{\cosh n} \right) = \pi r $$

(H. Ohtsuka)

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$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$


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 Post subject: Re: An infinite sum
PostPosted: Wed Apr 05, 2017 5:49 pm 

Joined: Tue May 10, 2016 3:56 pm
Posts: 30
Notice that
$$\frac{\sinh r}{\cosh n} = \frac{e^r - e^{-r}}{e^n + e^{-n}} = \frac{e^{r-n} - e^{-(r+n)}}{1 + e^{-2n}}.$$
Let $\alpha = e^{r-n}, \beta = e^{-(r+n)}$. Since $\alpha\beta = e^{-2n}$, by the formula
$$\arctan\alpha - \arctan\beta = \arctan\left(\frac{\alpha - \beta}{1 + \alpha\beta}\right),$$
we have
$$\frac{\sinh r}{\cosh n} = \arctan(e^{r-n}) - \arctan(e^{-(r+n)}).$$
Let $f(x) = \arctan(e^{-x})$. For positive integer $N > | r |$,
\begin{eqnarray}
\sum_{n=1}^N\,\left(\arctan(e^{r-n}) - \arctan(e^{-(r+n)})\right) & = & \sum_{n=1}^N\,(f(n-r) - f(n+r)) \nonumber\\
& = & \left(f(1-r) + f(2-r) + \cdots + f(r-1) + f(r)\right) \nonumber \\
&& - \left(f(N-r+1) + f(N-r + 2) + \cdots + f(N+r)\right).
\end{eqnarray}
Similarly, let $g(x) = \arctan(e^{x})$. For positive integer $N > | r |$,
\begin{eqnarray}
\sum_{n= -1}^{-N}\,\left(\arctan(e^{r-n}) - \arctan(e^{-(r+n)})\right) & = & \sum_{n=1}^N\,\left(\arctan(e^{r+n}) - \arctan(e^{n-r})\right)\nonumber \\
& = & \sum_{n=1}^N\,(g(n+r) - g(n-r)) \nonumber\\
& = & \left(g(N-r+1) + g(N-r + 2) + \cdots + g(N+r)\right) \nonumber \\
&& - \left(g(1-r) + g(2-r) + \cdots + g(r-1) + g(r)\right).
\end{eqnarray}
Since
\begin{eqnarray*}
\sum_{n = - N}^{N}\,\arctan\left(\frac{\sinh r}{\cosh n}\right) = \arctan(\sinh r) + \left\{\sum_{n=-1}^{-N}\, + \sum_{n=1}^{N}\right\}\,\left( \arctan(e^{r-n}) - \arctan(e^{-(r+n)})\right),
\end{eqnarray*}
in view of facts that $f(-x) = g(x)$, $\arctan(\sinh r) = \arctan(e^r) - \arctan(e^{-r}) = g(r) - f(r)$, by using (1) and (2), we find that
\begin{eqnarray}
\sum_{n = - N}^{N}\,\arctan\left(\frac{\sinh r}{\cosh n}\right) & = & \arctan(\sinh r) + \left(f(1-r) + f(2-r) + \cdots + f(r-1) + f(r)\right) \nonumber \\
&& - \left(f(N-r+1) + f(N-r + 2) + \cdots + f(N+r)\right) \nonumber\\
&& + \left(g(N-r+1) + g(N-r + 2) + \cdots + g(N+r)\right) \nonumber \\
&& - \left(g(1-r) + g(2-r) + \cdots + g(r-1) + g(r)\right)\nonumber\\
& = & - \left(f(N-r+1) + f(N-r + 2) + \cdots + f(N+r)\right) \nonumber\\
&& + \left(g(N-r+1) + g(N-r + 2) + \cdots + g(N+r)\right).
\end{eqnarray}
Notice that
\begin{eqnarray*}
\lim_{N \to \infty}\, g(N+k) & = & \lim_{N \to \infty}\,\arctan(e^{N+k}) = \frac{\pi}{2},\hspace{0.2in} \mbox{for all $-r + 1\leq k \leq r$},\\
\lim_{N \to \infty}\, f(N+k) & = & \lim_{N \to \infty}\,\arctan(e^{-(N+k)}) = 0,\hspace{0.2in} \mbox{for all $-r + 1\leq k \leq r$}.
\end{eqnarray*}
Letting $N \to \infty$ in (3) yields
$$\sum_{n = - \infty}^{\infty}\,\arctan\left(\frac{\sinh r}{\cosh n}\right) = 2r\cdot \frac{\pi}{2} = \pi r$$
as desired.


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 Post subject: Re: An infinite sum
PostPosted: Mon May 15, 2017 2:22 pm 

Joined: Sat Nov 14, 2015 6:32 am
Posts: 126
Location: Melbourne, Australia
Nice.. Thank you mathofusva! What do you say about this product:

$$\prod_{n=-\infty}^{\infty} \left( 1 + \frac{\sin ir}{\cosh n} \right) = e^{r^2 + \pi ir}$$

It does have an immediate connection with the above series.

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$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$


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