Is sequence $\sum_{k=1}^{n}\sin\left( k+\frac{1}{k}\right)$ bounded? If so, does $\sum_{k=1}^{\infty}\sin k-\sin\left(k+\frac{1}{k}\right)$ converge?

Question

It is well-known that the sequence $a_n:=\displaystyle\sum_{k=1}^{n} \sin k$ is bounded.

I want to see if $\displaystyle\sum_{k=1}^{\infty} \sin k - \sin\left( k+\frac{1}{k} \right)\ $ converges. The reason this is interesting is because $\sin k - \sin\left( k+\frac{1}{k} \right)\ $ is close to $\pm \frac{1}{k}$ quite often (for about half of the integers $k$).

But then I realised that I can't even prove that $b_n:=\displaystyle\sum_{k=1}^{n} \sin\left( k+\frac{1}{k} \right) $ is bounded, because the methods for proving that $\displaystyle\sum_{k=1}^{n} \sin k$ is bounded don't work for $\displaystyle\sum_{k=1}^{n} \sin\left( k+\frac{1}{k} \right). $ Let's see why:

$$\sum_{k=1}^{n} \sin\left( k+\frac{1}{k} \right) $$

$$ = \frac{\displaystyle\sum_{k=1}^n \left(e^{i\left( 1 + \frac{1}{k^2} \right)}\right)^k-\sum_{k=1}^n \left(e^{-i\left( 1 + \frac{1}{k^2} \right)}\right)^k}{2i} $$

But now we cannot say that $\displaystyle\sum_{k=1}^n z^k=z\frac{1-z^n}{1-z}\ $ where $\ z = e^{i\left( 1 + \frac{1}{k^2} \right)},$ because here $z$ is non-constant.

So my questions are:

1. Is my above analysis correct?

2. Is it known if $b_n:=\displaystyle\sum_{k=1}^{n} \sin\left( k+\frac{1}{k} \right) $ is bounded?

3. Does $\displaystyle\sum_{k=1}^{\infty} \sin k - \sin\left( k+\frac{1}{k} \right)\ $ converge?

Answer 1

First of all, your analysis seems correct: you can't use the same exact methods.

If you study $u_k = \sin(k) - \sin\left(k + 1/k\right)$, you easily get $u_k \sim -\sin'(k)/k = -\cos(k)/k$.

You could also have used (if you did not want to deal with derivative): $$\begin{align*}\sin k - \sin\left(k + 1/k\right) &= \sin k - \sin k \cos \frac{1}{k} - \cos k \sin \frac{1}{k} \\ &= \sin k \left(1-\cos\frac{1}{k}\right) - \cos k \sin \frac{1}{k}\\ &\sim \frac{\sin k}{2k^2} - \frac{\cos k}{k} \sim - \frac{\cos k}{k} \end{align*}$$

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Let us study $v_k = \frac{\cos k }{k}$. We put $S_n = \sum_{k=0}^n \cos k$. $$\begin{align*} \sum_{n=0}^N v_n &= \sum_{n=0}^N \frac{\cos n }{n} = \sum_{n=0}^N \frac{S_n - S_{n-1} }{n} = \sum_{n=0}^N \frac{S_n}{n} - \sum_{n=0}^N \frac{S_{n-1} }{n}\\ &= \sum_{n=0}^N \frac{S_n }{n} - \sum_{n=-1}^{N-1} \frac{S_n}{n+1} = \frac{S_N}{N} - S_{-1} + \sum_{n=0}^N S_n \left( \frac{1}{n} - \frac{1}{n+1}\right)\\ &= \frac{S_N}{N}- 0 +\sum_{n=0}^N \frac{S_n}{n(n+1)} \end{align*} $$ But since $S_n$ is bounded, this converges, hence $\sum u_k$ converges.

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NB: this trick with the $S_n$ can also be done with $R_n$ (series of the rest) and is called Abel's summation. This is a discrete version of integration by parts.

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Now obviously, this convergence plus the boundedness of $\sum \sin k$ implies the boundedness of $\sum \sin(k+1/k)$ since $\sin(k+1/k) = \sin k - u_k$.

Maybe you can try to show that if $\left(\sum_{k\leq n} a_k \right)_n$ is bounded and $\alpha>0$, then $\sum \frac{a_n}{n^\alpha}$ converges.