Show that $$\lim_{n \rightarrow \infty} (\cot(\frac{x}{n+1})-\cot(\frac{x}{n-1}))=\frac{2}{x}$$
One way to prove is using series but I am wondering is there any other way to prove it . Can anyone suggest me any ideas on how to solve it without series . Any help would be appreciated .
On
Not sure if this is what you want, but at least you don't have to use the entire Taylor series. If you are generous with the Big O notation it is pretty quick, $\cot(x) = \frac{1}{x+O(x^3)}$ as $x \to 0$ we have that
\begin{align*} \lim_{n \to \infty} \cot\left(\tfrac{x}{n+1}\right)-\cot\left(\tfrac{x}{n-1}\right)&= \lim_{n\to\infty}\left(\frac{1}{\frac{x}{n+1}+O(\tfrac{1}{n^3})} - \frac{1}{\frac{x}{n+1}+O(\tfrac{1}{n^3})} \right)\\ &= \lim_{n\to\infty}\left(\frac{n+1}{x+O(\tfrac{1}{n^2})} - \frac{n-1}{x+O(\tfrac{1}{n^2})} \right)\\ &= \frac{2}{x + \lim_{n\to \infty}O(1/n^2)} = \frac{2}{x}.\\ \end{align*}
On
$$\cot\left(\frac x{n+1}\right)-\cot\left(\frac x{n-1}\right)=\int_{\frac x{n+1}}^{\frac x{n-1}}\csc^2t\,dt\tag1$$ and for $x>0$ $$\frac{\frac {2x}{n^2-1}}{\sin^2(\frac x{n-1})}\leq \int_{\frac x{n+1}}^{\frac x{n-1}}\csc^2t\,dt\leq\frac{\frac {2x}{n^2-1}}{\sin^2(\frac x{n+1})}\tag2$$ By squeeze theorem, $(1)$ and $(2)$, we have $\lim_{n\to\infty}\cot\left(\frac x{n+1}\right)-\cot\left(\frac x{n-1}\right)=\frac 2x.$ Similarly for $x<0.$
On
If one has fun in trigonometry
$$\cot \left(\frac{x}{n+1}\right)-\cot \left(\frac{x}{n-1}\right)=\frac{\sin \left(\frac{x}{n-1}\right) \cos \left(\frac{x}{n+1}\right)-\sin \left(\frac{x}{n+1}\right) \cos \left(\frac{x}{n-1}\right)}{\sin \left(\frac{x}{n+1}\right) \sin \left(\frac{x}{n-1}\right)}=$$ $$\frac{\sin \left(\frac{2 x}{n^2-1}\right)}{\sin \left(\frac{x}{n+1}\right) \sin \left(\frac{x}{n-1}\right)}=\frac{2 \sin \left(\frac{x}{n^2-1}\right) \cos \left(\frac{x}{n^2-1}\right)}{\sin \left(\frac{x}{n+1}\right) \sin \left(\frac{x}{n-1}\right)} \to \frac{2 x}{\frac{\left(n^2-1\right) x^2}{n^2-1}}\to \frac{2}{x}$$
On
A little tedious but using the Digamma function also works. In particular, the $\psi$ function satisfies the reflection formula, $$\cot z=\frac{1}{\pi}\left(\psi\left(1-\frac{z}{\pi}\right)-\psi\left(\frac{z}{\pi}\right)\right)$$ we have, $$\cot\left(\frac{x}{n+1}\right)=\frac{1}{\pi}\left(\psi\left(1-\frac{x}{\pi(n+1)}\right)-\psi\left(\frac{x}{\pi(n+1)}\right)\right) \\ \cot\left(\frac{x}{n-1}\right)=\frac{1}{\pi}\left(\psi\left(1-\frac{x}{\pi(n-1)}\right)-\psi\left(\frac{x}{\pi(n-1)}\right)\right)$$ then, $$\cot\left(\frac{x}{n+1}\right)-\cot\left(\frac{x}{n-1}\right)=\frac{1}{\pi}(A-B)$$ where $$A=\psi\left(1-\frac{x}{\pi(n+1)}\right)-\psi\left(1-\frac{x}{\pi(n-1)}\right) \\ B=\psi\left(\frac{x}{\pi(n+1)}\right)-\psi\left(\frac{x}{\pi(n-1)}\right)$$ taking limits, $$\lim_{n\to\infty}A=\psi(1)-\psi(1)=0.$$ $B$ is a little more difficult as the $\psi$ function has a pole at zero, but by Taylor's formula, $$\psi(z+1)=-\gamma+\frac{\pi^2}{6}z+O(z^2)$$ as $z\to0$, and upon using the functional relation $\psi(z+1)=\psi(z)+1/z$ we find, $$\psi(z)=-\frac{1}{z}-\gamma+O(z)$$ as $z\to 0$, where $\gamma$ is the Euler–Mascheroni constant. Substitute, $$B=\frac{\pi(n-1)}{x}-\frac{\pi(n+1)}{x}+O\left(\frac{1}{n}\right)=-\frac{2}{x}+O\left(\frac{1}{n}\right)$$ as $n\to\infty$, hence, $$\lim_{n\to\infty}\frac{A-B}{\pi}=\frac{2}{x}.$$
As an intuition for this approach, note that this method seeks to remove the divergent $-1/z$ term in the Laurent expansion of $\psi(z)$.
I can show an intuitive proof: as $x \to 0$ $\tan(x)=x$ $$\lim_{n \to \infty} \cot(\frac{x}{n+1})-\cot(\frac{x}{n-1})=\lim_{n \to \infty} \frac{1}{\tan(\frac{x}{n+1})}-\frac{1}{\tan(\frac{x}{n-1})}$$ $$\cot(\frac{x}{n+1})-\cot(\frac{x}{n-1})=\lim_{n \to \infty} \frac{1}{\tan(\frac{x}{n+1})}-\frac{1}{\tan(\frac{x}{n-1})}=\lim_{n \to \infty} \frac{n+1}{x}-\frac{n-1}{x}=\lim_{n \to \infty} \frac{(n+1)-(n-1)}{x}=\frac{2}{x}$$ the actual proof involves a lot of $\epsilon$ and $\delta$ hope I helped!
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