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I have a function $$ g_N(x) =\frac{1}{x}+ \sum_{n=1}^\infty \Big(\frac{1}{x+n}+\frac{1}{x-n}\Big) $$ How can I prove that $$g_N(\frac{x}{2})+g_N(\frac{x+1}{2}) = 2g_N(x)\;?$$

Cocopuffs
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Supermartzin
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  • For $g_N(x/2)$, you have terms that look like $\frac{2}{x+2n}$, and for $g_N((x+1)/2))$ you have terms like $\frac{2}{x+2n+1}$. Notice that when you look at the sum from $-N$ to $N$, then the first terms are summing over even numbers, and for the second terms, you are summing over odd numbers.

    But, it seems like the right hand side should be something like $2g_{2N+1}(x)$, so could you double check the problem statement?

     – Alex R. Apr 09 '13 at 23:22
  • This should be full problem statement. – Supermartzin Apr 09 '13 at 23:33
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    It no longer depends on $N$. It is now just $\pi \cdot \mathrm{cot}(\pi x)$. – Cocopuffs Apr 09 '13 at 23:35
  • If $N$ is infinity, then my comment carries through. – Alex R. Apr 09 '13 at 23:40

2 Answers2

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Using $(7)$ from this answer, we get $$ g_N(x)=\pi\cot(\pi x) $$ Therefore, we have $$ \begin{align} g_N\left(\frac{x}{2}\right)+g_N\left(\frac{x+1}{2}\right) &=\pi\cot\left(\pi\frac{x}{2}\right)+\pi\cot\left(\pi\frac{x+1}{2}\right)\\ &=\pi\cot\left(\pi\frac{x}{2}\right)-\pi\tan\left(\pi\frac{x}{2}\right)\\ &=2\pi\frac{1-\tan^2\left(\pi\frac{x}{2}\right)}{2\tan\left(\pi\frac{x}{2}\right)}\\ &=2\pi\cot(\pi x)\\[15pt] &=2g_N(x) \end{align} $$

Community
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robjohn
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$$g_N(\frac{x}{2}) =\frac{2}{x}+ \sum_{n=1}^\infty \Big(\frac{2}{x+2n}+\frac{2}{x-2n}\Big)$$ $$g_N(\frac{x+1}{2}) =\frac{2}{x+1}+ \sum_{n=1}^\infty \Big(\frac{2}{x+2n+1}+\frac{2}{x-2n+1}\Big)$$

Adding them together you get

$$g_N(\frac{x}{2})+g_N(\frac{x+1}{2})=\frac{2}{x}+\frac{2}{x+1}+\sum_{n=1}^\infty \Big(\frac{2}{x+2n}+\frac{2}{x-2n}\Big)+\sum_{n=1}^\infty \Big(\frac{2}{x+2n+1}+\frac{2}{x-2n+1}\Big)$$

$$=\frac{2}{x}+\frac{2}{x+1}+\sum_{n=1}^\infty \Big(\frac{2}{x+2n}+\frac{2}{x-2n}\Big)+\sum_{n=1}^\infty \Big(\frac{2}{x+2n+1}\Big)+\Big( \sum_{n=1}^\infty \frac{2}{x-2n+1}\Big)$$ $$=\frac{2}{x}+\frac{2}{x+1}+\sum_{n=1, even}^\infty \Big(\frac{2}{x+n}+\frac{2}{x-n}\Big)+\sum_{n=3, odd}^\infty \Big(\frac{2}{x+n}\Big)+\Big( \sum_{n=1, odd}^\infty \frac{2}{x-n}\Big)$$

Group $\frac{2}{x+1}$ with the second sum and you are done.

P.S. Just be careful, the series are not absolutely convergent, so to do the actual computations you need to work with partial sums. The partial sums on left and right have different "end points".

N. S.
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