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I have to show that the series $\displaystyle{\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}\sin(1+\frac{x}{n})}$ converges uniformly in $[-a,a], a>0$. $$$$ That's what I have tried:

$w_n=\frac{(-1)^n}{\sqrt{n}} \sin1 \cos(\frac{x}{n})+\frac{(-1)^n}{\sqrt{n}} \cos1 \sin(\frac{x}{n})=\frac{(-1)^n}{\sqrt{n}} \sin1 (\cos(\frac{x}{n})+1-1)+\frac{(-1)^n}{\sqrt{n}} \cos1 \sin(\frac{x}{n})= \frac{(-1)^n}{\sqrt{n}} \sin1 (\cos(\frac{x}{n})-1)+\frac{(-1)^n}{\sqrt{n}} \sin1 +\frac{(-1)^n}{\sqrt{n}} \cos1 \sin(\frac{x}{n})=\sin1 \cdot g_n(x)+ \cos1 \cdot f_n(x)+\frac{(-1)^n}{\sqrt{n}} \sin1 $ .

We set $d_n(x)=w_n-\frac{(-1)^n}{\sqrt{n}} \sin1=\sin1 \cdot g_n(x)+ \cos1 \cdot f_n(x)$

$|g_n(x)|=|\frac{(-1)^n}{\sqrt{n}}(\cos(\frac{x}{n})-1)|=\frac{1}{\sqrt{n}} (\cos(\frac{x}{n})-1) (\cos(\frac{x}{n})+1)=\frac{1}{\sqrt{n}}(\sin^2(\frac{x}{n})) \leq \frac{1}{\sqrt{n}} \cdot \frac{x^2}{n^2} \leq \frac{1}{\sqrt{n}} \cdot \frac{a^2}{n^2}=\frac{a^2}{n^{\frac{5}{2}}}$

$\sum_{n=1}^{\infty} \frac{a^2}{n^{\frac{5}{2}}}< +\infty$.

$|f_n(x)|=|\frac{(-1)^n}{\sqrt{n}} \sin(\frac{x}{n})| \leq \frac{1}{\sqrt{n}} \frac{x}{n} \leq \frac{a}{n^{\frac{3}{2}}}$

$\sum_{n=1}^{\infty} \frac{a}{n^{\frac{3}{2}}}< +\infty$.

Also,from Dirichlet's test,we know that $\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}$ converges,let $b:=\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}$

Therefore,we have $d_n(x) \leq \sin1 \frac{a^2}{n^{\frac{5}{2}}}+ \cos1 \frac{a}{n^{\frac{3}{2}}}$.

$\sum_{n=1}^{\infty} \sin1 \frac{a^2}{n^{\frac{5}{2}}}+ \cos1 \frac{a}{n^{\frac{3}{2}}}$ converges,as $\sum_{n=1}^{\infty} \frac{a^2}{n^{\frac{5}{2}}}$ converges and $\sum_{n=1}^{\infty} \frac{a}{n^{\frac{3}{2}}}$ converges too.

So, $\sum_{n=1}^{\infty} d_n(x)$ converges uniformly to a $s(x)$. $\sum_{n=1}^{\infty} d_n(x)=\sum_{n=1}^{\infty} (w_n-\frac{(-1)^n}{\sqrt{n}} \sin1)= \sum_{n=1}^{\infty} w_n-b \overset{uniformly}{=}s(x) \Rightarrow \sum_{n=1}^{\infty} w_n=b+s(x):=q(x):[-a,a] $

So,the series converges uniformly to $q(x)$.

Could you tell me if it is right?

evinda
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1 Answers1

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Hint You may use Dirichlet's criterion, since $\sin (xn^{-1})$ is eventually decreasing in $[0,M]$ and $1-\cos (xn^{-1})$ is eventually decreasing in $[0,M]$ too, and both go uniformly to zero, and $$\sum_{n\geqslant 1} (-1)^{n}n^{-1/2}$$ converges.

Pedro
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  • I haven't understood it.. Do I use Dirichlet's criterion,because $b_n=\frac{1}{\sqrt{n}} \to 0$ and is decreasing and the partial sum $s=\sum_{k=1}^{n} (-1)^k \sin1 \cos(\frac{x}{k})$ is bounded??If yes, why is the partial sum bounded? – evinda May 11 '14 at 13:52
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    @evinda Re-read what I wrote. Dirichlet's criterion wants a decreasing sequence $a_n(x)$ that goes monotonically to zero, and a sequence of functions whose sum is bounded, $b_n(x)$. I took $a_n(x)=\sin(xn^{-1}),1-\cos(xn^{-1})$ and $b_n(x)=(-1)^n n^{-1/2}$. – Pedro May 11 '14 at 13:55
  • I edited my post...Could you tell me if it is better now? – evinda May 11 '14 at 15:17
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    I cannot understand why this was downvoted. – Pedro Dec 05 '14 at 19:25
  • @PedroTamaroff Excellent proof. Don't worry i upvoted it – Magneto Jul 11 '18 at 14:17