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How to show that $\exists M, \forall n: |\overset{n}{\underset{k = 1}{\sum}}\cos{(k + \frac{1}{k})}| \leq M$?

I tried to prove by finding real part of $\overset{n}{\underset{k = 1}{\sum}}e^{i(k + \frac{1}{k})}$, but it didn't work out.

Also since $|\overset{n}{\underset{k = 1}{\sum}}\cos{(k + \frac{1}{k})}| = |\overset{n}{\underset{k = 1}{\sum}}(\cos{k}\cos{\frac{1}{k}} - \sin{k}\sin{\frac{1}{k}})| \leq |\overset{n}{\underset{k = 1}{\sum}}\cos{k}\cos{\frac{1}{k}}| + |\overset{n}{\underset{k = 1}{\sum}}\sin{k}\sin{\frac{1}{k}}| $, I tried to find upper bounds for each module, but unsuccessfully.

mokrota21
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    Find a bound on $\vert\sum_{k=1}^n\cos k\vert$, and use that $\cos\frac1k\to1$ (write for example $\cos\frac1k=1+\varepsilon_k$ with $\varepsilon_k\to0$). Do the same with the sinus part, with $\sin\frac1k\to0$. – Will Oct 29 '22 at 09:46
  • @Will I'm not really good at calculus can you please explain why in $|\overset{n}{\underset{k = 1}{\sum}}\cos{k}\cos{\frac{1}{k}}|$ we can change $\cos{\frac{1}{k}}$ with 1 because it is limit? No offense I just can't intuitively justify this. – mokrota21 Oct 29 '22 at 10:01
  • Nvm just realized why it is true. Thank you – mokrota21 Oct 29 '22 at 10:09
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    Can you show us what you did? The steps you need to do from the initial start I proposed do not appear that trivial to me. In particular you might need to do something in the spirit of Dirichlet's test: https://en.wikipedia.org/wiki/Dirichlet%27s_test. In particular, the thing I have in mind use the fact that $cos\frac1k-1$ and $sin\frac1k$ are both monotonous. – Will Oct 29 '22 at 10:17
  • @Will Bound: $|\overset{n}{\underset{k = 1}{\sum}}\cos{k}| \leq \frac{1}{|\sin{\frac{1}{2}}|}$. About why we can change $\cos{\frac{1}{k}}$ with 1 I thought that since $\overset{n}{\underset{k = 1}{\sum}}\cos{k} - \overset{n}{\underset{k = 1}{\sum}}\cos{k}\cos{\frac{1}{k}} = \overset{n}{\underset{k = 1}{\sum}}\cos{k}(1 - \cos{\frac{1}{k}})$, we can say that for some m big enough $\overset{n}{\underset{k = m}{\sum}}\cos{k}(1 - \cos{\frac{1}{k}}) \to 0$ and the rest is finite sum – mokrota21 Oct 29 '22 at 10:30
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    "and the rest is finite sum". You mean $\sum_{k=1}^m\cos k(1-\cos\frac1k)$? In your proof you don't fix $m$, you let it be any integer large enough, so it is not valid. – Will Oct 29 '22 at 10:34

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We have $$\cos(k+k^{-1})=\cos k\cos (k^{-1})-\sin k\sin(k^{-1})\\ =\cos k-(1-\cos(k^{-1}))\cos k-\sin k\sin(k^{-1})$$ The partial sums $\sum_{k=0}^n\cos k$ and $\sum_{k=1}^n\sin k$ are bounded due to the explicit formula $$\sum_{k=0}^n(\cos k+i\sin k)=\sum_{k=0}^n e^{ik}={e^{i(n+1)}-1\over e^i-1}$$ Thus series $$\sum_{k=1}^\infty [1-\cos(k^{-1})]\cos k,\quad \sum_{k=1}^\infty \sin (k^{-1})\sin k$$ are convergent by the Dirichlet test, as the sequences $1-\cos (k^{-1})$ and $\sin(k^{-1})$ are decreasing and convergent to $0.$ Therefore the partial sums of these series are bounded. Summarizing the sums $$\sum_{k=0}^n\cos(k+k^{-1})$$ are bounded since they are represented by three sums, each of them bounded.

Ryszard Szwarc
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It is known that $\sum_{k=1}^n\cos{k}$ and $\sum_{k=1}^n\sin{k}$ are bounded by $$\Big|\sum_{k=1}^{n} (e^{i})^k\Big|\le \frac2{|1-e^{i}|}=\frac{2}{\sqrt{(1-\cos(1))^2+\sin^2(1)}}.$$ Moreover, $\{\sin\frac{1}{k}\}_{k\geq 1}$ and $\{1-\cos\frac{1}{k}\}_{k\geq 1}$ decrease monotonically to $0$. Hence, by Dirichlet's test, the following series are convergent $$\sum_{k=1}^{\infty}\cos{k}(1-\cos\frac{1}{k}) \quad \text{and}\quad\sum_{k=1}^{\infty}\sin{k}\sin\frac{1}{k}.$$ So we may conclude that the right-hand side of $$\begin{align}\Big|\sum_{k=1}^n\cos{(k + \frac{1}{k})}\Big| &= \Big|\sum_{k=1}^n(\cos{k}\cos{\frac{1}{k}} - \sin{k}\sin{\frac{1}{k}})\Big|\\ &\leq \Big|\sum_{k=1}^n\cos{k}\Big|+\Big|\sum_{k=1}^n\cos{k}(1-\cos{\frac{1}{k}})\Big| + \Big|\sum_{k=1}^n\sin{k}\sin{\frac{1}{k}}\Big| \end{align}$$ is bounded.

Using summation by parts, we find an explicit upper bound: $$\Big|\sum_{k=1}^n\cos{(k + \frac{1}{k})}\Big|\leq \frac{2(1+2(1-\cos(1))+2\sin(1))}{\sqrt{(1-\cos(1))^2+\sin^2(1)}}.$$

Robert Z
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Here's my attempt.

Theorem. (Dirichlet's test). If monotone $a_n$ satisfy$\lim\limits_{n \to \infty} a_n = 0$ and $\sum_\limits{k=1}^n b_k$ is bounded, then $\sum\limits_{k=1}^na_kb_k$ exists.

We need a bound for $|\overset{n}{\underset{k = 1}{\sum}}\cos{k}\cos{\frac{1}{k}}| + |\overset{n}{\underset{k = 1}{\sum}}\sin{k}\sin{\frac{1}{k}}|$

Note that $\cos \frac{1}{k}-1$ is monotone and has limit $0$, $$ |\overset{n}{\underset{k = 1}{\sum}}\cos{k}(\cos{\frac{1}{k}}-1)|$$ exists by the Dirichlet test and thus bounded; $ |\overset{n}{\underset{k = 1}{\sum}}\cos{k}|$ itself is also bounded as one should verify, and hence there's a bound to the first term.

Similarly since $|\sum\limits_{k=1}^n \sin k|$ is bounded and $\sin \frac{1}{k}$ is monotone and has limit $0$, we can again apply Dirichlet's test and conclude $|\overset{n}{\underset{k = 1}{\sum}}\sin{k}\sin{\frac{1}{k}}|$ converges and hence has a bound too.

Lab
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