Prove $\sum\limits_{n = 1}^\infty \frac{( - 1)^n}{\ln n + \sin n}$ is convergent.

Help prove the alternating series $$\sum\limits_{n = 1}^\infty \frac{(-1)^n}{\ln n + \sin n}$$ is convergent. $$\frac 1 {\ln n + \sin n}$$ is a decreasing sequence but it is not motonically decreasing. I am not sure how to deal with this situation.

My failed attempt..

For even terms,

$$\sum_{n = 1}^\infty \frac{( - 1)^{2n}}{\ln 2n + 1} \le \sum_{n = 1}^\infty \frac{( - 1)^{2n}}{\ln 2n + \sin 2n} \leqslant \sum_{n = 1}^\infty \frac{( - 1)^{2n}}{\ln 2n - 1}$$

where the two "bound" series do not converge

For odd terms

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

where the two "bound" series do not converge.

• Try using the fact that $-1\leq \sin(n)\leq 1$. – DMcMor Apr 5 '17 at 0:01
• The function $\frac1{\ln x+\sin x}$ is not eventually decreasing. – Greg Martin Apr 5 '17 at 0:09
• First time I've seen "{\text{ + }}" in MathJax code where simply "+" was appropriate. – Michael Hardy Apr 5 '17 at 0:35
• This is pretty hardcore. I don't think this will be any easier than math.stackexchange.com/questions/21175/… (which is the same series, only with $\sin(n)$ replaced by $\cos(n)$). – levap Apr 5 '17 at 0:52
• In the @levap's link, the series with $\cos(n)$ converges. The same method applies to this one. Thus, this series converges. – Sungjin Kim Apr 5 '17 at 23:17

Does $\sum\limits_{n = 1}^\infty \frac{(-1)^n}{\ln n + \sin n}$ converge or diverge?

I think that it probably diverges, but I can show that the similar sum $\sum\limits_{n = 1}^\infty \frac{(-1)^n}{\ln n + (-1)^n}$ diverges and $\sum\limits_{n = 1}^\infty \frac{(-1)^n}{\ln n + c}$ converges for any real $c$ such that $\ln(n)+c \ne 0$ for all $n$ (in particular for $c > 0$).

If $f(n)$ is bounded, let $s_m =\sum\limits_{n = 1}^m \frac{(-1)^n}{\ln n + f(n)}$ and let $s =\lim_{m \to \infty} s_m$ if the limit exists.

$\begin{array}\\ s_{2m} &=\sum\limits_{n = 1}^{2m} \frac{(-1)^n}{\ln n + f(n)}\\ &=\sum\limits_{n = 1}^{m} (\frac{-1}{\ln (2n-1) + f(2n-1)}+\frac{1}{\ln (2n) + f(2n)})\\ &=\sum\limits_{n = 1}^{m} \frac{-(\ln (2n) + f(2n))+(\ln (2n-1) + f(2n-1))}{(\ln (2n-1) + f(2n-1))(\ln (2n) + f(2n))}\\ &=\sum\limits_{n = 1}^{m} \frac{\ln (1-1/(2n)) + f(2n-1)-f(2n)}{(\ln (2n-1) + f(2n-1))(\ln (2n) + f(2n))}\\ &=\sum\limits_{n = 1}^{m} \frac{\ln (1-1/(2n))}{(\ln (2n-1) + f(2n-1))(\ln (2n) + f(2n))}+\sum\limits_{n = 1}^{2m} \frac{f(2n-1)-f(2n)}{(\ln (2n-1) + f(2n-1))(\ln (2n) + f(2n))}\\ &=u_m+v_m\\ \end{array}$

where $u_m =\sum\limits_{n = 1}^{2m} \frac{\ln (1-1/(2n))}{(\ln (2n-1) + f(2n-1))(\ln (2n) + f(2n))}$ and $v_m =\sum\limits_{n = 1}^{2m} \frac{f(2n-1)-f(2n)}{(\ln (2n-1) + f(2n-1))(\ln (2n) + f(2n))}$.

Each term in $u_m$ is, for large enoough $n$, less in absolute value than $\frac1{n\ln^2(n)}$ (because $\ln(1-1/(2n)) \approx -\frac1{2n}$ and $(\ln (2n-1) + f(2n-1))(\ln (2n) + f(2n)) \approx \frac1{\ln^2(n)}$) and the sum of these converges.

Therefore convergence depends on $v_m$. Each term in $v_m$ is $\frac{f(2n-1)-f(2n)}{(\ln (2n-1) + f(2n-1))(\ln (2n) + f(2n))} \approx \frac{f(2n-1)-f(2n)}{\ln^2(2n)}$.

If $f(n) = c$, where $c$ is a constant, then $f(2n-1)-f(2n) = 0$, so $v_m = 0$ and the sum converges.

If $f(n) = (-1)^n$, then $f(2n-1)-f(2n) = (-1)-(1) = -2$ and the sum of $\frac{-2}{\ln^2(n)}$ diverges.

If $f(n) = \sin(n)$,

$\begin{array}\\ f(2n-1)-f(2n) &=\sin(2n-1)-\sin(2n)\\ &=2\sin((2n-1-2n)/2)\cos((2n-1+2n)/2)\\ &=2\sin(-1/2)\cos(2n-1/2)\\ \end{array}$

so the sum depends on the sum $\sum_{n=1}^m \frac{\cos(2n-1/2)}{\ln^2(2n)}$.

According to Wolfy, $\sum_{n=1}^m \cos(2 n - 1/2) = \csc(1) \sin(m) \cos(m + 1/2)$ so this sum is bounded but not convergent. I think that this implies that $\sum_{n=1}^m \frac{\cos(2n-1/2)}{\ln^2(2n)}$ does not converge. This might be proved using a variation of summation by parts, but I am not sure how to do this, so I will leave it at this.

• Good job, I was starting to suspect this sum diverges, just because nobody can manage to prove it converges. But I'd still love to see a complete rigorous proof. – Gregory Grant Jul 26 '17 at 20:14
• Your sums after the first one should be to $m$ not to $2m$. Also how do you get the bound on $u_m$ as less than $\frac1{n\ln^2(n)}$ for large enough $n$? – Gregory Grant Jul 26 '17 at 23:30
• Thanks. I made the correction and added an additional explanation. – marty cohen Jul 27 '17 at 4:04
• The integral $$\int_1^\infty \frac{\cos(2t-1/2)}{\ln^2(2t)} \,dt$$ does converge (use integration by parts, differentiating $\frac1{\ln^2(2t)}$ and integrating $\cos(2t-1/2)$). So I suspect the sum $$\sum_{n=1}^\infty \frac{\cos(2n-1/2)}{\ln^2(2n)}$$ also converges, and indeed summation by parts is probably how it can be proved. – Greg Martin Jul 27 '17 at 4:31
• What I consider probable is that the sum is bounded but oscillates, so that it does not converge to a limit. – marty cohen Jul 27 '17 at 4:43