Limit and a series $$\lim_{n\to+\infty}⁡\frac{\cos⁡(\frac1n)+\cos⁡(\frac2n)+⋯+\cos⁡(\frac nn)}{n}$$
What I have tried:
I tried rewriting the series portion as the sum $i=1$ to n of $\cos(i/n)$. Then used the formula the sum $i=1$ to n of i = n(n+1)/2 and substituted for i. Then i had the lim(n→∞)⁡[(1/n)cos(n(n+1)/(2n)}] which would be just 0


Applying Bounds and the Squeeze Theorem

Aside from Riemann sums, we can evaluate the limit of interest by noting that the cosine function, $\cos(x)$ is monotonically decreasing and positive for $0 \le x\le 1$.  Hence, we have
$$\frac1n \int_1^{n+1} \cos(x/n)\,dx\le \frac1n\sum_{k=1}^n\cos(k/n)\le \frac1n \cos(1)+\frac1n \int_1^n \cos(x/n)\,dx \tag 1$$
Carrying out the integrals in $(1)$ we find that 
$$\sin(1+1/n)-\sin(1/n)\le \frac1n\sum_{k=1}^n\cos(k/n)\le \frac1n \cos(1)+\sin(1)-\sin(1/n)$$
whereupon application of the squeeze theorem yield the coveted result

$$\lim_{n\to \infty}\frac1n\sum_{k=1}^n\cos(k/n)=\sin(1)$$

