2
$\begingroup$

$$\lim_{n\to+\infty}⁡\frac{\cos⁡(\frac1n)+\cos⁡(\frac2n)+⋯+\cos⁡(\frac nn)}{n}$$

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

  • 2
    You seem to be implying that $\cos(x)+\cos(y)=\cos(x+y)$, but that is definitely not the case. As an alternative: can you recognize this as a Riemann sum for an integral? – 2017-02-03
  • 0
    The result is $\sin(1)$ as can be shown using two ingredients: 1) as $\cos(x) = Re( \exp( i x))$ your sum is geometric. 2) $\exp( i /n) - 1 \sim i/n$ for $n \to \infty$ – 2017-02-04

2 Answers 2

4

We have $$\lim_{n\to \infty} \frac{1}{n}\sum_{j=1}^n \cos \frac{j}{n}= \int_0^1 \cos x \;\mathrm d x = \sin 1.$$

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

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

  • 0
    Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark – 2017-02-15