How do I show that the partial sums of the sequence $a_n=\cos( \log n)$ are bounded?

Question

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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?

Answer 1

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.

Answer 2

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.

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