Calculate $\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=1}^{n}k\sin\left(\frac{a}{k}\right)$ I'm trying to calculate $\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=1}^{n}k\sin\left(\frac{a}{k}\right)$. Intuitively the answer is $a$, but I can't see any way to show this. Can anyone help? Thanks!
 A: $\displaystyle \sin x \leq x $ for $x \geq 0.$ Integrating this over $[0,t]$ gives $$ -\cos t +1 \leq \frac{t^2}{2} . $$
Integrating both sides again from $[0,x]$ gives $$ -\sin x + x \leq \frac{x^3}{6} .$$ 
Thus, $$ x - \frac{x^3}{6} \leq \sin x \leq x.$$ 
Hence, $$ \frac{1}{n} \sum_{k=1}^{n} k \left( \frac{a}{k} - \frac{a^3}{6k^3} \right ) \leq 

 \frac{1}{n} \sum_{k=1}^{n} k \sin \left( \frac{a}{j} \right) \leq  \frac{1}{n} \sum_{k=1}^{n} k \left( \frac{a}{k} \right). $$
Since $ \displaystyle \sum_{k=1}^n \frac{a^3}{6k^3} $ is convergent, the Squeeze theorem shows that $\displaystyle  \frac{1}{n} \sum_{k=1}^{n} k \sin \left( \frac{a}{j} \right) \to a.$
A: I’m not going to work out all of the details; rather, I’ll suggest in some detail a way to approach the problem.
First, it suffices to prove the result for $a>0$, since the sine is an odd function. For $a>0$ we have $k\sin\left(\frac{a}k\right)<k\left(\frac{a}k\right)=a$, so $$\frac1n\sum_{k=1}^nk\sin\left(\frac{a}k\right)<\frac1n\sum_{k=1}^na=a\;;$$ this gives you an upper bound of $a$ on any possible limit.
You know that $\lim\limits_{x\to 0}\frac{\sin x}x=1$, so there is a $c>0$ such that $\sin x>\frac{x}2$ whenever $0<x<c$. This means that $$k\sin\left(\frac{a}k\right)>\frac{a}2$$ whenever $\frac{a}k<c$, i.e., whenever $k>\frac{a}c$. Now suppose that $n$ is very large compared with $\frac{a}c$; then ‘most’ of the terms of $$\frac1n\sum_{k=1}^nk\sin\left(\frac{a}k\right)\tag{1}$$ will be greater than $\frac{a}2$, and hence so will $(1)$ itself. You may have to do a little fiddling to say just how big $n$ should be taken relative to $\frac{a}c$, but it should be clear that this idea works to show that the limit of $(1)$ as $n\to\infty$ must be at least $\frac{a}2$.
But what I did with $\frac12$ can clearly be done with any positive fraction less than $1$: if $0<\epsilon<1$, there is a $c>0$ such that $\sin x>\epsilon x$ whenever $0<x<c$. If you’ve filled in the missing details for the previous paragraph, you shouldn’t have too much trouble generalizing to show that the limit of $(1)$ must be at least $\epsilon a$ for any $\epsilon <1$ and hence must be at least $a$.
