Squeeze me

Using the inequality $|\sin x - x| \le \tfrac16x^3$ for $0 \le x \le 1$ , we see that $\left| \sum_{k=1}^n \sin \left(\frac{\pi}{\sqrt{n^2+k}}\right) -\sum_{k=1}^n \frac{\pi}{\sqrt{n^2+k}}\right| \;\le \; \sum_{k=1}^n \frac{\pi^3}{6(n^2 + k)^{\frac32}} \; \le \; \frac{\pi^3}{6n^2} \;,$ for $n \ge 3$ , and so we want to evaluate $S \; = \; \lim_{n\to\infty}\sum_{k=1}^n \sin\left(\frac{\pi}{\sqrt{n^2+k}}\right) \; = \; \lim_{n\to\infty}\pi\sum_{k=1}^n \frac{1}{\sqrt{n^2+k}} \;.$ Now $\sqrt{\frac{n}{n+1}} \; = \; \frac{n}{\sqrt{n^2+n}} \; \le \; \displaystyle\sum_{k=1}^{n} \frac{1}{\sqrt{n^2+k}} \; \le \;\frac{n}{\sqrt{n^2+1}} \; < \; 1$ Since $\frac{n}{n+1}$ converges to $1$ as $n \to \infty$ , the Squeeze Theorem can now be applied to give $S = \pi$ .

It is always risky to discard a limit of a sum of an infinite number of terms just because the limit of each term is zero, just in case the sum of those terms does not converge to zero.

Using Taylor series for sine, we have:

$\begin{aligned} \lim_{n \to \infty} \sum_{k=1}^n \sin \left( \frac{\pi}{\sqrt{n^2+k}} \right) & = \lim_{n \to \infty} \sum_{k=1}^n \left[\left( \frac{\pi}{\sqrt{n^2+k}} \right) - \frac{1}{3!}\left( \frac{\pi}{\sqrt{n^2+k}} \right)^3 + \frac{1}{5!}\left( \frac{\pi}{\sqrt{n^2+k}} \right)^5-... \right] \\ & = \lim_{n \to \infty} \sum_{k=1}^n \frac{\pi}{\sqrt{n^2+k}} \quad \quad \small \color{#3D99F6}{\text{As } \frac{\pi}{\sqrt{n^2+k}} \to 0 \text{, later terms are negligible (see Note). }} \\ & = \sum_{k=1}^n \frac{\pi}{n} = n \times \frac{\pi}{n} = \pi \approx \boxed{3.142} \end{aligned}$

$\color{#3D99F6}{\text{Note:}}$

$\text{As } \frac{\pi}{\sqrt{n^2+k}} \to 0 \quad \Rightarrow \frac{\pi}{\sqrt{n^2+k}} \gg \left(\frac{\pi}{\sqrt{n^2+k}}\right)^3 \gg \left(\frac{\pi}{\sqrt{n^2+k}}\right)^5 \gg ...$

$\large\lim_{n\to\infty}\sum_{k=1}^n \sin\left(\frac{\pi}{\sqrt{n^2+n+0.25}}\right)<\lim_{n\to\infty}\sum_{k=1}^n \sin\left(\frac{\pi}{\sqrt{n^2+k}}\right)<\lim_{n\to\infty}\sum_{k=1}^n \sin\left(\frac{\pi}{\sqrt{n^2+0}}\right)$

$\large\lim_{n\to\infty}n\sin\left(\frac{\pi}{n+0.5}\right)<\lim_{n\to\infty}\sum_{k=1}^n \sin\left(\frac{\pi}{\sqrt{n^2+k}}\right)<\lim_{n\to\infty}n\sin\left(\frac{\pi}{n}\right)$

The left limit:

$\displaystyle\large\lim_{n\to\infty}n\sin\left(\frac{\pi}{n+0.5}\right)=\large\lim_{n\to\infty}\left(\frac{\sin\left(\frac{\pi}{n+0.5}\right)}{\frac{\pi}{n+0.5}}\right)\left(\frac{n}{n+0.5}\right)(\pi)=1\times1\times\pi=\pi$

The right limit:

$\displaystyle\large\lim_{n\to\infty}n\sin\left(\frac{\pi}n\right)=\large\lim_{n\to\infty}\left(\frac{\sin\left(\frac{\pi}n\right)}{\frac{\pi}n}\right)(\pi)=1\times\pi=\pi$

The limit is $\pi$ by the Squeeze theorem.