How to show $\exists M, \forall n: |\overset{n}{\underset{k = 1}{\sum}}\cos{(k + \frac{1}{k})}| \leq M$? 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.
 A: 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.
A: 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:
$$\Big|\sum_{k=1}^n\cos{(k + \frac{1}{k})}\Big|\leq
\frac{2(1+2(1-\cos(1))+2\sin(1))}{\sqrt{(1-\cos(1))^2+\sin^2(1)}}.$$
A: 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.
