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Examine the convergence of the series of functions $$\displaystyle\mathop{\sum}\limits_{n=1}^{+\infty}\Big({\frac{x}{1+x^n}}\Big)^n$$ a) pointwise in $[0,1]$,

b) uniformly in $[0,1]$.

My attempt for pointwise convergence: For all $x\in[0,1)$ exists $n_0(x)\in{\mathbb{N}}$ such that for all $n\in\mathbb{N}$ with $n\geqslant n_0(x)$ : $$\displaystyle\Big|\Big({\frac{x}{1+x^n}}\Big)^n\Big|<\frac{1}{n^2}\,.$$ Because $\sum_{n=1}^{+\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$, we have that the series $\sum_{n=1}^{+\infty}\big({\frac{x}{1+x^n}}\big)^n$ converges pointwise in $[0,1)$. Also for $x=1$ : $\sum_{n=1}^{+\infty}\big({\frac{1}{1+1^n}}\big)^n=\sum_{n=1}^{+\infty}\big({\frac{1}{2}}\big)^n=1$. So, the series converges pointwise in $[0,1]$.

I have no answer for uniform convergence.

edit: This is not an answer for the uniform convergence issue. I'm just giving two plots which shows the behavior of the partial sums sequence $S_n=\sum_{k=1}^{n}\big({\frac{x}{1+x^k}}\big)^k$ near $1$, where is possible the non-uniform convergence of the series $\sum_{n=1}^{+\infty}\big({\frac{x}{1+x^n}}\big)^n$, for helping others to procced further. In the rest of the interval the series looks that converges uniformly. enter image description here

enter image description here

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  • $\begingroup$ Hint: Optimze the function $x\mapsto \frac{x}{1+x^{n}}$ on [0,1] $\endgroup$ Jan 8 '18 at 20:11
  • $\begingroup$ @TheOscillator The sequence of functions $f_n(x):=\frac{x}{1+x^n}\,,\; n\in\mathbb{N}\,,$ does not converges uniformly on $[0,1]$. But how can we use that for the uniform convergence of the series $\sum_{n=1}^{+\infty}\big(f_n(x)\big)^n$ ? There exists a big gap between these two. $\endgroup$ Jan 9 '18 at 13:13
  • $\begingroup$ That’s interesting ! I don’t see how this question is off topic !! $\endgroup$
    – Tolaso
    Jan 13 '18 at 15:45
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Hint:

$$\sqrt[n]{\left(\frac x{1+x^n}\right)^n} =\frac x{1+x^n}\xrightarrow[n\to\infty]{}\begin{cases}x,&x\in[0,1)\\{}\\ \cfrac12,&x=1\end{cases}$$

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  • $\begingroup$ Sorry, but I don't understand why this hint tell us something about uniform convergence. $\endgroup$ Jan 8 '18 at 20:44
  • $\begingroup$ For pointwise convergence: For all $x\in[0,1)$ exists $n_0(x)\in{\mathbb{N}}$, such that for all $n\in{\mathbb{N}}$ with $n\geqslant n_0(x)$ : $$\displaystyle\Big|\Big({\frac{x}{1+x^n}}\Big)^n\Big|<\frac{1}{n^2}\,.$$ Because $\sum_{n=1}^{+\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$, we have that the series $\sum_{n=1}^{+\infty}\big({\frac{x}{1+x^n}}\big)^n$ converges pointwise in $[0,1)$. Also $\sum_{n=1}^{+\infty}\big({\frac{1}{1+1^n}}\big)^n=\sum_{n=1}^{+\infty}\big({\frac{1}{2}}\big)^n=1$. So, the series converges point wise in $[0,1]$. $\endgroup$ Jan 8 '18 at 20:47
  • $\begingroup$ @GrigoriosKostakos The hint tells you there's pointwise convergence if $\;0\le x<1\;$ , but divergence if $\;x=1\;$ ...This already tells you stuff about u.c., doesn't it? $\endgroup$
    – DonAntonio
    Jan 8 '18 at 22:33
  • $\begingroup$ No, does not. As stated above the series converges at $x=1$ because $\sum_{n=1}^{+\infty}\big({\frac{1}{1+1^n}}\big)^n=\sum_{n=1}^{+\infty}\big({\frac{1}{2}}\big)^n=1$. Does not diverges. So, uniform convergence remains open. $\endgroup$ Jan 9 '18 at 3:58
  • $\begingroup$ @GrigoriosKostakos Good point, it is true. Then the OP will still need to make an evaluation as he tried to do from the beginning...and in fact that'll be his answer. $\endgroup$
    – DonAntonio
    Jan 9 '18 at 8:16

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