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Problem: Find the following limit $${\displaystyle \lim_{n\rightarrow\infty}\left[\left(1+\frac{1}{n}\right)\sin\frac{\pi}{n^{2}}+\left(1+\frac{2}{n}\right)\sin\frac{2\pi}{n^{2}}+\ldots+\left(1+\frac{n-1}{n}\right)\sin\frac{\left(n-1\right)}{n^{2}}\pi\right]}$$.

Attempt: Observe that \begin{align*} & \lim_{n\rightarrow\infty}\left[\left(1+\frac{1}{n}\right)\sin\frac{\pi}{n^{2}}+\left(1+\frac{2}{n}\right)\sin\frac{2\pi}{n^{2}}+\ldots+\left(1+\frac{n-1}{n}\right)\sin\frac{\left(n-1\right)}{n^{2}}\pi\right]\\ = & \lim_{n\rightarrow\infty}\left[\left(1+\frac{1}{n}\right)\left(\frac{\pi}{n^{2}}+O\left(\frac{\pi^{3}}{n^{6}}\right)\right)+\ldots+\left(1+\frac{n-1}{n}\right)\left(\left(\frac{n-1}{n^{2}}\right)\pi+O\left(\frac{\left(n-1\right)^{3}\pi^{3}}{n^{6}}\right)\right)\right]\\ = & \lim_{n\rightarrow\infty}\left[\left(\sum_{k=1}^{n}\left(1+\frac{k}{n}\right)\frac{k\pi}{n^{2}}\right)+\left(\sum_{k=1}^{n}\left(1+\frac{k}{n}\right)O\left(\frac{k^{3}\pi^{3}}{n^{6}}\right)\right)\right]\\ = & \lim_{n\rightarrow\infty}\left[\left(\sum_{k=1}^{n}\left(1+\frac{k}{n}\right)\frac{k\pi}{n^{2}}\right)+\sum_{k=1}^{n}\left(1+\frac{k}{n}\right)O\left(\frac{1}{n^{3}}\right)\right]\tag{1} \end{align*} Note that \begin{align*} \lim_{n\rightarrow\infty}\sum_{k=1}^{n}\left(1+\frac{k}{n}\right)\frac{1}{n^{3}} & =\lim_{n\rightarrow\infty}\left(\frac{1}{n^{2}}+\sum_{k=1}^{n}\frac{k}{n^{4}}\right)\leq\lim_{k\rightarrow\infty}\left(\frac{1}{n^{2}}+\frac{1}{n^{3}}\right)=0. \end{align*} Hence \begin{align*} \left(1\right) & =\lim_{n\rightarrow\infty}\left[\sum_{k=1}^{n}\left(1+\frac{k}{n}\right)\frac{k\pi}{n^{2}}\right]=\pi\lim_{n\rightarrow\infty}\left[\sum_{k=1}^{n}\left(\frac{k}{n}+\left(\frac{k}{n}\right)^{2}\right)\frac{1}{n}\right]\\ & =\pi\int_{0}^{1}x+x^{2}dx=\frac{5\pi}{6}. \end{align*}

Question I don't know if I did it correctly.

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  • $\begingroup$ And your question is ...? $\endgroup$ – PhoemueX Mar 13 '16 at 6:49
  • $\begingroup$ I am not sure that I see the question. Could you clarify ? $\endgroup$ – Claude Leibovici Mar 13 '16 at 6:49
  • $\begingroup$ Is this a proof verification? $\endgroup$ – Henricus V. Mar 13 '16 at 6:52
  • $\begingroup$ You have the answer correct. It is $\frac{5\pi}{6}$$ Hmm this link doesn't paste correctly - use copy and paste to get it all: wolframalpha.com/input/?i=Limit%5BSum%5B(1+%2B+i%2Fn)+Sin%5Bi*Pi%2Fn%5E2%5D,+%7Bi,+1,+n+-+1%7D%5D,+n+-%3E+Infinity%5D $\endgroup$ – Ian Miller Mar 13 '16 at 7:27
  • $\begingroup$ What is even funny is that, for finite values of $n$, the summation has a closed form. $\endgroup$ – Claude Leibovici Mar 13 '16 at 7:31
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Yes, you did this perfectly except for one (inconsequential) mistake. In "note that", you should have

$$ \begin{align*} \sum_{k=1}^n \left( 1 + \dfrac{k}{n} \right) \dfrac{1}{n^3} & = \dfrac{1}{n^2} + \sum_{k=1}^n \dfrac{k}{n^4} \\ & < \dfrac{1}{n^2} + \mathbf{ \sum_{k=1}^n \dfrac{n}{n^4} } \\ & = \dfrac{1}{n^2} + \dfrac{1}{n^2} \end{align*} $$

which still tends to 0.

If you'd like some empirical confidence boosting,

Mathematica produces this convergence on a numerical level very clearly.
(source: pages.iu.edu)

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