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Let $\{f_n\}$ be a sequence of functions that converge uniformly to $f$ in $(0,1]$. For every sequence $\{x_n\}$ such that $x_n\rightarrow0^+$, the series $\sum_{n=1}^\infty f(x_n)$ diverges. Show that $\sum f_n(x_n)$ also diverges for every sequence $x_n\rightarrow0^+$.

My try-

By negation, we assume there exists a sequence $\{y_n\}$, such that $y_n\rightarrow 0^+$ and $\sum f_n(y_n)$ converges to a function $S(y_n)$.

I want to show that $\sum f(y_n)$ also converges, yielding a contradiction. So we have

$|\sum f(y_n)-S(y_n)|\leq|\sum f_n(y_n)-\sum f(y_n)|+|\sum f_n(y_n)-S(y_n)|$

$|\sum f_n(y_n)-S(y_n)|$ is as small as we want because of the convergence, and I know $|f_n(y_n)-f(y_n)|$ is as small as we want thanks to uniform convergence.

But what about $|\sum f_n(y_n)-\sum f(y_n)|$?

Does $f_n\rightarrow f$ uniformly suggest $\sum f_n\rightarrow\sum f$ uniformly?

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  • $\begingroup$ One thing that doesn't seem right is you say $\sum f_n(y_n)$ converges to a function, but that function would be constant since it is an infinite sum of values, not functions. $\endgroup$ – Yeldarbskich Oct 3 '15 at 18:47
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Let $L = \liminf_{x\to 0^+} |f(x)|.$ If $L = 0,$ then there is a sequence $x_n \to 0^+$ such that $|f(x_k)| \to 0.$ This implies there is a subsequence $n_k$ such that $|f(x_{n_k})|\le 1/2^k.$ Hence the series $\sum f(x_{n_k})$ converges absolutely. This is a contradiction since $x_{n_k} \to 0^+.$

So we know $L>0.$ Thus there is $b>0$ such that $|f(x)| > L/2$ for $x\in (0,b).$ To finish, use the uniform convergence of $f_n$ to $f$ to show there exists $N$ such that for all $n\ge N,$ $|f_n(x)|> L/4$ for all $x\in (0,b).$ This leads to the desired conclusion.

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