What is the partial sum formula for $\sum\limits_{j=2}^{x} \csc^2\left(\frac{\pi}{j}\right)$?

Question

I want to find the partial sum formula for: $$ \sum_{j=2}^{x} \csc^2\left(\frac{\pi}{j}\right) $$

I found a similar one, but I don't how to use it here:

Finite Sum $\sum\limits_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}$

If I try to find approximation by the Fundamental Theorem of Engineering, $$\sin\left(x\right)\approx x$$

Then I get, $$ \sum_{j=2}^{x} \csc^2\left(\frac{\pi}{j}\right) = \sum_{j=2}^{x} \frac{1}{\sin^{2}\left(\frac{\pi}{j}\right)} \approx \sum_{j=2}^{x} \frac{1}{\left(\frac{\pi}{j}\right)^2} $$

Which gives in WolframAlpha:

$$ \sum_{j=2}^{x} \frac{1}{\left(\frac{\pi}{j}\right)^2} = \frac{2x^3 + 3x^2 + x - 6}{6\pi^2} $$

And the graph of it was so close to the original sum that it blew my mind.

Answer 1

I shall derive the asymptotics of the sum for large $n$. Differentiation of the Laurent series of $\cot(\pi t)$ yields $$ \csc ^2 (\pi t) = \frac{1}{{\pi ^2 }}\frac{1}{{t^2 }} + \frac{1}{3} + \frac{2}{{\pi ^2 }}\sum\limits_{k = 1}^\infty (2k + 1)\zeta (2k + 2)t^{2k} , $$ for $|t|<\pi$, where $\zeta$ is the Riemann zeta function. Thus, \begin{align*} &\sum\limits_{j = 2}^n {\csc ^2 \left( {\frac{\pi }{j}} \right)} = \frac{1}{{\pi ^2 }}\sum\limits_{j = 2}^n {j^2 } + \frac{{n - 1}}{3} + \frac{2}{{\pi ^2 }}\sum\limits_{k = 1}^\infty {(2k + 1)\zeta (2k + 2)\sum\limits_{j = 2}^n {\frac{1}{{j^{2k} }}} } \\ & = \frac{1}{{3\pi ^2 }}n^3 + \frac{1}{{2\pi ^2 }}n^2 + \frac{{2\pi ^2 + 1}}{{6\pi ^2 }}n - \frac{{\pi ^2 + 3}}{{3\pi ^2 }} + \frac{2}{{\pi ^2 }}\sum\limits_{k = 1}^\infty {(2k + 1)\zeta (2k + 2)\sum\limits_{j = 2}^n {\frac{1}{{j^{2k} }}} } \\ & = \frac{1}{{3\pi ^2 }}n^3 + \frac{1}{{2\pi ^2 }}n^2 + \frac{{2\pi ^2 + 1}}{{6\pi ^2 }}n + C - \frac{2}{{\pi ^2 }}\sum\limits_{k = 1}^\infty {(2k + 1)\zeta (2k + 2)\zeta (2k,n + 1)} , \end{align*} for any $n\ge 2$, where $$ C = - \frac{{\pi ^2 + 3}}{{3\pi ^2 }} + \frac{2}{{\pi ^2 }}\sum\limits_{k = 1}^\infty {(2k + 1)\zeta (2k + 2)(\zeta (2k) - 1)} = 0.10982248806823565160 \ldots $$ and $\zeta(s,a)$ is the Hurwitz zeta function. Employing the known integral formula for this function, we find that $$ \sum\limits_{j = 2}^n {\csc ^2 \left( {\frac{\pi }{j}} \right)} = \frac{1}{{3\pi ^2 }}n^3 + \frac{1}{{2\pi ^2 }}n^2 + \frac{{2\pi ^2 + 1}}{{6\pi ^2 }}n + C+ \int_0^{ + \infty } {{\rm e}^{ - (n+1/2)t} F(t)\,{\rm d}t} $$ for any $n\ge 2$, where $$ F(t) = - \frac{2}{{\pi ^2 }}\frac{{t/2}}{{\sinh (t/2)}}\sum\limits_{k = 0}^\infty {(2k + 3)\zeta (2k + 4)\frac{{t^{2k} }}{{(2k + 1)!}}} $$ for $t>0$. An application of Watson's lemma then shows that \begin{multline*} \sum\limits_{j = 2}^n {\csc ^2 \left( {\frac{\pi }{j}} \right)} \sim \frac{1}{{3\pi ^2 }}n^3 + \frac{1}{{2\pi ^2 }}n^2 + \frac{{2\pi ^2 + 1}}{{6\pi ^2 }}n + C - \frac{{2\pi ^2 }}{{15}}\frac{1}{{2n + 1}} \\ - \frac{{2\pi ^2 (40\pi ^2 - 63)}}{{2835}}\frac{1}{{(2n + 1)^3 }} - \frac{{2\pi ^2 (336\pi ^4 - 2000\pi ^2 + 2205)}}{{70875}}\frac{1}{{(2n + 1)^5 }} + \ldots \end{multline*} as $n\to+\infty$. Setting $m=2n+1$, we can write \begin{multline*} \sum\limits_{j = 2}^n {\csc ^2 \left( {\frac{\pi }{j}} \right)} \sim \frac{1}{{24\pi ^2 }}m^3 + \frac{{4\pi ^2 - 1}}{{24\pi ^2 }}m + C' - \frac{{2\pi ^2 }}{{15}}\frac{1}{m} \\ - \frac{{2\pi ^2 (40\pi ^2 - 63)}}{{2835}}\frac{1}{{m^3 }} - \frac{{2\pi ^2 (336\pi ^4 - 2000\pi ^2 + 2205)}}{{70875}}\frac{1}{{m^5 }} + \ldots \end{multline*} as $n\to+\infty$, with $$ C' = - \frac{{\pi ^2 + 2}}{{2\pi ^2 }} \!+\! \frac{2}{{\pi ^2 }}\!\sum\limits_{k = 1}^\infty {(2k + 1)\zeta (2k + 2)(\zeta (2k) - 1)} = - 0.05684417859843101506 \ldots \,. $$