How do I show that the partial sums of the sequence $a_n=\cos( \log n)$ are bounded? 
I am just stuck. How do I show that the partial sums of the sequence $a_n=\cos( \log n)$ are bounded? Log n makes everything twisted...Could anyone please help me?
 A: It's not bounded. Take some $k$, take partial sum from $a_k = \lceil \exp(2 \pi k - \pi / 4)\rceil$ to $b_k = \lfloor \exp(2 \pi k + \pi /4)\rfloor$.
For $n \in [a_k, b_k]$ we have $\log n \in [2 \pi k -\pi/4; 2\pi k + \pi / 4]$, so $\cos(\log n) \geqslant \frac{\sqrt 2}{2}$.
So each term is at least $\frac{\sqrt 2}{2}$, and there are at least $$b_k - a_ k \geqslant\\ \exp(2 \pi k + \pi / 4) - \exp(2 \pi k - \pi / 4) - 2 =\\ \exp(2 \pi k - \pi / 4) \cdot (\exp(\pi) - 1) - 2$$ of them terms.
So partial sums over some segments can be arbitrarily large, thus sequence of partial sums isn't bounded.
A: Intuitive explanation:
For large $n$, the sum converges to an integral, i.e. the area between the curve and the axis $x$.
In the successive intervals $[e^{2n\pi},e^{2(n+1)\pi}]=e^{2n\pi}\cdot[1,e^{2\pi}]$, the function takes the same values while the interval length grows exponentially.
Hence,
$$\int_1^{e^{2(n+1)\pi}}\cos(\log x)\,dx=\sum_{k=0}^n e^{2k\pi}\int_1^{e^{2\pi}}\cos(\log x)\,dx.$$
So unless the integral is zero, which is not true as the curve is asymmetric, the sum diverges.

The argument is not rigorous because we replaced the sum by an integral, but the error goes decreasing with larger $n$.
