Finding $\lim_{n\to\infty}\sum_{k=1}^n \sin(\frac\pi{\sqrt{n^2+k}})$

Question

Find $$\lim_{n\to\infty}\sum_{k=1}^n \sin(\frac\pi{\sqrt{n^2+k}})$$

I think this has to be evaluated with riemann sums. I tried using $\sin x\approx x$ as n is big, but that got me to $\int_0^1 (1+1/x)^{-1/2}$ but that gives me the wrong answer. Any suggestions?

EDIT: I would like to see a solution with Riemann sums.

Answer 1

I think this can be done without Riemann sums.

You may use the fact that, as $n \to \infty$, $$ \sin\left(\frac\pi{\sqrt{n^2+k}}\right) =\frac\pi{\sqrt{n^2+k}}+O\left(\frac\pi{(n^2+k)^{3/2}}\right) $$ to get $$ \sum_{k=1}^n\sin\left(\frac\pi{\sqrt{n^2+k}}\right) =\sum_{k=1}^n\frac\pi{\sqrt{n^2+k}}+\sum_{k=1}^nO\left(\frac\pi{(n^2+k)^{3/2}}\right). $$ On the one hand, as $n \to \infty$, $$ \frac{\pi\: n}{\sqrt{n^2+n}}\leq \sum_{k=1}^n\frac\pi{\sqrt{n^2+k}}\leq\frac{\pi\: n}{\sqrt{n^2+1}} $$ giving, as $n \to \infty$,

$$ \sum_{k=1}^n\frac\pi{\sqrt{n^2+k}} \to \pi. $$

On the other hand, there exists $C\geq 0$ such that, as $n \to \infty$, $$ \frac{C\pi\: n}{(n^2+n)^{3/2}}\leq \sum_{k=1}^nO\left(\frac\pi{(n^2+k)^{3/2}}\right)\leq\frac{C\pi\: n}{(n^2+1)^{3/2}} $$ giving, as $n \to \infty$,

$$ \sum_{k=1}^nO\left(\frac\pi{(n^2+k)^{3/2}}\right) \to 0. $$

Finally,

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

Answer 2

Use the inequality $$x - \frac{1}{6}x^3 \leq \sin x \leq x, \quad x \in [0, \pi/2).$$ For sufficiently large $n$, the summand satisfies: $$\frac{\pi}{\sqrt{n^2 + k}} - \frac{1}{6}\frac{\pi^3}{(n^2 + k)^{3/2}} \leq \sin\left(\frac{\pi}{\sqrt{n^2 + k}}\right) \leq \frac{\pi}{\sqrt{n^2 + k}} \tag{1}.$$ For $k \in \{1, \ldots, n\}$, the right side bound in $(1)$ is bounded above by $\frac{\pi}{n}$, while the left side bound in $(1)$ is bounded below by $$\frac{\pi}{\sqrt{n^2 + n}} - \frac{1}{6}\frac{\pi^3}{n^3}.$$ It thus follows that $$\frac{n\pi}{\sqrt{n^2 + n}} - \frac{1}{6}\frac{\pi^3}{n^2} \leq \sum_{k = 1}^n\sin\left(\frac{\pi}{\sqrt{n^2 + k}}\right) \leq \pi.$$ By the squeeze principle, we conclude that $$\lim_{n \to \infty} \sum_{k = 1}^n\sin\left(\frac{\pi}{\sqrt{n^2 + k}}\right) = \pi.$$