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As was mentioned in the title, can we say that both the series $$ \sum{\frac{\sin (n+1/n)}{\ln n}} \quad \text{ and } \quad \sum{\frac{\sin n}{\ln n}} $$ converges or diverges simultaneously? I mean, it looks very intuitive since $$ \sin(n+1/n) \approx \sin(n) $$ for sufficiently large $n$. But I didn't find any rigorous Test or Rule to show that.

Levon Minasian
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  • It's not obvious to me that $\sin(n+1/n) \approx \sin(n)$ for sufficiently large $n.$ Are you sure this is true? If so, where is a proof? I have not found one. – Adam Rubinson Feb 28 '21 at 10:07
  • For the last one see the proof here https://math.stackexchange.com/questions/342637/does-sum-frac-sin-nn-converge?noredirect=1&lq=1 on davidlouwrydouda's solution. You can use the same Dirichlet criterion to show it converges. – Vlad Matei Feb 28 '21 at 10:11
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    @AdamRubinson $$\begin{align*} \sin!\left(n+\frac{1}{n}\right) &= \sin n \cos \frac{1}{n} + \cos n \sin \frac{1}{n} \ &= \sin n\left(1-\frac{1}{n^2} + o!\left(\frac{1}{n^2}\right) \right)
    • \cos n\left(\frac{1}{n} + o!\left(\frac{1}{n^2}\right) \right) \&= \sin n + O(1/n)

    \end{align*}$$

     – Clement C. Feb 28 '21 at 10:11
  • @VladMatei But the question is not to decide whether the second converges; it is whether one can argue that both have same nature. – Clement C. Feb 28 '21 at 10:11
  • @ClementC. yeah, that's all correct. – Levon Minasian Feb 28 '21 at 10:12
  • Well with the asymptotic expansions you wrote the proof is done by the same dirichlet criterion. You can just bound the tails – Vlad Matei Feb 28 '21 at 10:12
  • @VladMatei See my answer. There is some work to be done :) – Clement C. Feb 28 '21 at 10:13
  • They both converge in short and you have essentially proved that below. – Vlad Matei Feb 28 '21 at 10:15
  • @VladMatei They both converge, but proving that they have the same nature involves handling $\sum_n \frac{\cos n}{n\ln n}$ (which is basically as hard as proving convergence of $\sum_n \frac{\sin n}{\ln n}$). So it doesn't feel like proving "they both have same nature" is any simpler than proving "they both are convergent." – Clement C. Feb 28 '21 at 10:18

1 Answers1

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Since $\sin(a+b) = \sin a \cos b + \cos a \sin b$, by a Taylor expansion around $0$ we get $$\begin{align*} \sin\!\left(n+\frac{1}{n}\right) &= \sin n \cos \frac{1}{n} + \cos n \sin \frac{1}{n} \\ &= \sin n\left(1-\frac{1}{n^2} + o\!\left(\frac{1}{n^2}\right) \right) + \cos n\left(\frac{1}{n} + o\!\left(\frac{1}{n^2}\right) \right) \end{align*}$$ which means that $$\begin{align*} \frac{\sin\!\left(n+\frac{1}{n}\right)}{\ln n} &= \frac{\sin n}{\ln n} + \frac{\cos n}{n\ln n}+ o\!\left(\frac{1}{n^2}\right) \end{align*}$$ since $|\sin|,|\cos|\leq 1$. This shows that, if you take for granted that $$ \sum_{n=2}^\infty \frac{\cos n}{n\ln n} $$ converges, then indeed $$ \sum_{n=2}^\infty \frac{\sin\!\left(n+\frac{1}{n}\right)}{\ln n},\quad\sum_{n=2}^\infty \frac{\sin n}{\ln n} $$ have same nature (since the remainder, "$\sum_n o\!\left(\frac{1}{n^2}\right)$", is absolutely convergent by comparison). The question then becomes: can one easily show that $$ \sum_{n=2}^\infty \frac{\cos n}{n\ln n} $$ converges? (One way: Dirichlet's test.)

Clement C.
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