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

I am asked to prove that $\sum v_n$ converges uniformly.

This is my attempt: Let $$S_n(x)= \sum_{k=0}^{n}v_n$$ and $$F(x)=\sum_{k=0}^{\infty}v_n$$

I have to show that $|F(x)-S_n(x)|$ converges. But: $$|F(x)-S_n(x)|= \sum^{\infty}_{k=n+1}\frac{1}{2\sqrt{k}+cos(x)}$$

And this is where I am stuck. By looking at this sum, I would say that it diverges yet I am asked to prove it converges. Any help would be appreciated.

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    Your last statement is wrong, that is $$|F(x)-S_n(x)|=\left| \sum^{\infty}_{k=n+1}\frac{(-1)^k}{2\sqrt{k}+\cos(x)}\right|\neq\sum^{\infty}_{k=n+1}\frac{1}{2\sqrt{k}+\cos(x)}$$ – 2017-02-20
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    @Masacroso Indeed, I made an error. Based on this, I can now use the Alternating series test to show that it converges? – 2017-02-20
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    Do you know Dirichlet's criterion? – 2017-02-20
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    @JuliánAguirre I have not seen it in the class. – 2017-02-20
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    And Leibniz's criterion for alternating series? – 2017-02-20
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    @JuliánAguirre This one I do know. – 2017-02-20

3 Answers 3

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Point wise convergence follows from Leibniz's criterion. For each real number $x$ $$ \frac{1}{2\sqrt{n}+ \cos(x)}>0, $$ decreases as a function of $n$ and converges to $0$.

To prove uniform convergence you need the uniform version of the criterion: $$ \frac{1}{2\sqrt{n}+ \cos(x)} $$ decreases as a function of $n$ and converges uniformly to $0$. You can prove it adapting the proof of Leibniz's criterion to this case.

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Note that $$\frac{1}{2\sqrt{n}+\cos x}-\frac{1}{2\sqrt{n+1}+\cos x}=2\frac{\sqrt{n+1}-\sqrt{n}}{(2\sqrt{n}+\cos x)(2\sqrt{n+1}+\cos x)}$$ $$=\frac{2}{(\sqrt{n+1}+\sqrt{n})(2\sqrt{n}+\cos x)(2\sqrt{n+1}+\cos x)}\leq \frac{c}{(\sqrt{n})^3}$$

for some value of $c$.

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    Why are you subtracting the next term? – 2017-02-20
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    Its alternating isnt it ? – 2017-02-20
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Hint: $|F(x)-S_n(x)|=|\sum^{\infty}_{k=n+1}\frac{(-1)^k}{2\sqrt{k}+cos(x)}| \le |\sum^{\infty}_{k=n+1}\frac{(-1)^k}{2\sqrt{k}-1}| \le |\sum^{\infty}_{k=n+1}\frac{(-1)^k}{\sqrt{k}-1}|$.

Show that the series $\sum^{\infty}_{k=1}\frac{(-1)^k}{\sqrt{k}-1}$ is convergent and use that $N(\epsilon)$ in above inequality to get $|F(x)-S_n(x)| \lt \epsilon \;$ for $\; n \gt N(\epsilon)$.