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Is the following series of functions uniformly convergent on $\mathbb{R}$?

$$\sum_{n=2}^{\infty} \frac{(-1)^{n+1}} {\sqrt n + \cos x}$$

My attempt: My answer is No

I know that by Leibnitz test the given series converges, but by using $M$- test

$$\sup_{x \in \mathbb{R}}\left|\sum_{n=2}^{\infty} \frac{(-1)^{n+1}} {\sqrt n + \cos x}\right| \le \sup_{x \in \mathbb{R}} \left|\sum\frac{1}{\sqrt n}\right| \neq 0$$

so the given series is not uniformly convergent on $\mathbb{R}$

Is it correct ?

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  • $\begingroup$ Why is the inequality after "My attempt" true? If it were true, how would it allow you to answer the question? $\endgroup$ – Did Sep 23 '18 at 9:14
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Yes, it converges uniformly on $\mathbb R$. You just apply Dirichlet's test for uniform convergence:

  1. $\sum_{n=0}^N(-1)^n$ is uniformly bounded;
  2. for each real $x$, $\left(\dfrac1{\sqrt n+\cos(x)}\right)_{n\in\mathbb N}$ is monotonic;
  3. $\left(\dfrac1{\sqrt n+\cos(x)}\right)_{n\in\mathbb N}$ converges uniformly to $0$.
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Simply use the usual bound on the remainder of an alternating series $$\left|\sum_{k=n}^{\infty} \frac{(-1)^{k+1}} {\sqrt k + \cos x}\right|\leq \frac{1}{\sqrt n +\cos(x)}\leq \frac{1}{\sqrt n-1}$$

Since the bound is independent of $x$ and goes to $0$, there is indeed uniform convergence.

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