TITLE: Calculate $\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=1}^{n}k\sin\left(\frac{a}{k}\right)$
QUESTION [8 upvotes]: 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!

REPLY [3 votes]: $\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 $\displaystyle  \frac{1}{n} \sum_{k=1}^{n} k \sin \left( \frac{a}{j} \right) \to a.$