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.


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.
