Uniform convergence of $\sum_{n= 2}^{\infty}\dfrac{(-1)^n}{n+\sin(2\pi x)} $ I want to show $$\sum_{n = 2}^\infty\frac{(-1)^n}{n＋\sin(2\pi x)} $$ uniformly converges.
I tried to apply M-test, but by simply taking absolute value, $$\frac{1}{n+\sin(2\pi x)}$$
diverges.
So I tried to estimate the series more rigorously, like Starting from the first two terms, add then in pairs, then we can get $\dfrac{1}{n＋1＋\sin(2\pi x)}- \dfrac{1}{n＋\sin(2\pi x)}$ as majorant and this has square term in denominator, so we can apply M-test. But this is wrong because
$\dfrac{1}{n＋1＋\sin(2\pi x)}$ does not absolute converge, so I'm at a loss. I want to give a strict proof to this question.Thank you for your help.
 A: Yours is a convergent  alternating series. Say $s(x)=\sum^\infty_{n=2} \frac{(-1)^n}{n+\sin(2\pi x)}$, and $s_n$ is the n-th partial sum. Then, by well know property of alternating series
$|s(x)-s_n(x)|\leq \frac{1}{n+1+\sin(2\pi x)}\leq \frac{1}{n}$
The rest should be straight forward

Here is the property of alternating series, and a sketch of its proof,  that is useful to remember as this situation appears frequently in analysis:
Theorem: Suppose $a_n\searrow0$ as $n\rightarrow 0$ and let $s_n=\sum^n_{k=1}(-1)^{k+1}a_k$. Then $s_n$ converges so some $s\in\mathbb{R}$ and
$$0<(-1)^n(s-s_n)\leq a_{n+1}$$

*

*Convergence can be establish by the $M$-test that you quoted.

*As for the approximation part, notice
$$(-1)^n(s-s_n) = \sum_{k\geq1}(-1)^{k+1}a_{n+k}=\sum_{k=1}(a+2k-1-a_{n+2k})>0$$
and
$$
(-1)^n(s_n-a)=a_{n+1}-\sum_{k\geq1}(a_{n+2k}-a_{n+2k+1})\leq a_{n+1}
$$
As a consequence of all this,
$$|s-s_n|\leq a_{n+1}$$
A: The proof is given by Dirichlet's test for uniform convergence.
Let $f_n(x)=(-1)^n $ and $g_n(x)=\frac{1}{n+\sin{2 \pi x}}$.
Clearly, partial sums of sequence $\{f_n\}$ are unifomrly bounded, the sequence $\{g_n(x)\}$ is decreasing for all $x\in \mathbb{R}$, and $\{g_n\}$ converges pointwise to zero.
Now, $\frac {1}{n+\sin{2\pi x}} \leq \frac{1}{n-1}$, and therefore convergence of $\{g_n\}$ is uniform
Conclusion follows by Dirichlet's test
A: $$\sum_{i = 2}^{∞}\dfrac{（-1）^n}{n＋\sin（2\pi x）} = \sum_{i = 2}^{∞} \frac{（-1）^n}{n}\frac{1}{1+\frac{\sin（2 \pi x）}{n}}$$
Here $\sum_{i = 2}^{∞} \frac{（-1）^n}{n}$ converged and $\frac{1}{1+\frac{\sin（2 \pi x）}{n}}$ is bounded and monotonic, so we can use Abel test
There is also uniform convergence test.
