Convergence of $\sum_{n=1}^{\infty} \sin\big((-1)^n\log(1+n^{-1})+n^{-2}\big)$

Question

Study the convergence of teh following series:

$$\sum_{n=1}^{\infty} \sin\bigg((-1)^n\log\bigg(1+\frac{1}{n}\bigg)+\frac{1}{n^2}\bigg)$$

Using the fact that $\sin(-x)=-\sin(x)$, I wrote again the series in the form:

$$\sum_{n=1}^{\infty} (-1)^n\sin\bigg(\log\bigg(1+\frac{1}{n}\bigg)+\frac{(-1)^n}{n^2}\bigg)$$

Then using Taylor series: $$\sum_{n=1}^{\infty} (-1)^n\bigg(\frac{1}{n}-\frac{1}{2n^2}+o\bigg(\frac{1}{n^2}\bigg)+\frac{(-1)^n}{n^2}\bigg)$$

My aim would be that of using Leibniz, but how to handle $o$ in this case? Thank you.

Answer 1

Note that $$\sin\bigg((-1)^n\log\bigg(1+\frac{1}{n}\bigg)+\frac{1}{n^2}\bigg)=(-1)^na_n+b_n$$ where $$a_n:=\sin\bigg(\log\bigg(1+\frac{1}{n}\bigg)\bigg)\cos\bigg(\frac{1}{n^2}\bigg),\quad b_n:=\cos\bigg(\log\bigg(1+\frac{1}{n}\bigg)\bigg)\sin\bigg(\frac{1}{n^2}\bigg)$$ Show that $x\to \sin(\log(1+x))\cos(x^2)$ is positive and increasing in a right neighbourhood of zero. Hence $a_n$ decreases to zero and $\sum_{n}(-1)^na_n$ is convergent by Leibniz. On the other hand $\sum_{n}b_n$ is convergent because $b_n\sim 1/n^2$. Hence the given series is convergent.

Answer 2

Let $a_n:= \sin\bigg(\log\bigg(1+\frac{1}{n}\bigg)+\frac{1}{n^2}\bigg)$. Then $(a_n)$ is decreasing and $a_n \to 0$. Hence , by Leibniz (not Leibnitz !):

the series is convergent.

Answer 3

$|sin (x +y)-sin y|\leq |x|$ by MVT. Since $ \sum\sin (\frac 1 {n^{2}})$ is absoluetly convergent (because $|\sin x | \leq |x|$) it suffices to consider the series withous the $\frac 1 {n^{2}}$ term. Here you can pull out the $(-1)^{n}$ and use the theorem that $\sum (-1)^{n} a_n$ is convergent of $a_n$ decreases to $0$.