Prove the series $\sum_{n=1}^{\infty}\frac{(-1)^n}{\ln{n}+\sin{n}}$ converge How to prove this serie
$$\sum_{n=1}^{\infty}\frac{(-1)^n}{\ln{n}+\sin{n}}$$
converge? I can't do a comparison test with the Leibniz formula for $\pi$ because the series are not $>0$ for all $n$. I can't do a ratio test because I can't compute the limit, the alternating series test can't be applied, the absolute serie is not convergent. I'm out of ideas.
Any clues?
 A: Let $$a_n = \frac{(-1)^n}{\sin n + \ln n }$$
We wish to show that $\sum a_n$ converges. 
In order to do this, first note that $a_n$ is negative when $n$ is odd, and positive when $n$ is even. 
We will write $\sum a_n < \sum b_i + \sum c_j$, where $b_i$ are negative terms (with only odd indices $i$) that are smaller in absolute value than the negative terms $a_i$, and $c_j$ are positive terms (with only even indices $j$) that are larger in absolute value than the positive terms $a_j$. 

We want to choose $\{b_i\}$ and $\{c_j\}$ to satisfy the following conditions:


*

*$\sum b_i$ (sum taken over odd $i$) converges by the alternating series test

*$\sum c_j$ (sum taken over even $j$) converges by the alternating series test

*$|a_i| > |b_i|$ or equivalently, $a_i < b_i$, for all odd $i$

*$a_j < c_j$ for all even $j$

For $i$ odd, let $$b_i = \frac{-1}{2 + \ln n } > a_i$$
For $j$ even, let $$c_j = \frac{1}{-2 + \ln n } > a_i$$
(I am choosing $2$ and $-2$ here because they are greater than the maximum of $\sin n$ and less than the minimum of $\sin n$, respectively)

Note that $b_i$ and $c_j$ are both monotonically decreasing for $i,j > 10$.
(I am choosing $10$ here because it is greater than $e^2$, to avoid negative denominators due to $-2 + \ln n$)  
Therefore, by the alternating series test, we know that the following sums must converge:
$$\displaystyle\sum_{i=11,\, i \text{ odd}}^\infty b_i$$
$$\displaystyle\sum_{j=10,\, j \text{ even}}^\infty c_j$$

Therefore, $\sum a_n$ is bounded above by the sum of two convergent series: $$\displaystyle\sum_{n=10}^\infty a_n < \left(\displaystyle\sum_{i=11,\, i \text{ odd}}^\infty b_i \right) + \left(\displaystyle\sum_{j=10,\, j \text{ even}}^\infty c_j\right)$$
