Is there a sequence $f_{n}:\mathbb{R} \to \mathbb{R}$ such that the series $\sum \limits_{k=1}^{\infty}f_{n}(x)$ converges uniformly to $f(x)$, but $\sum \limits_{k=1}^{\infty}|f_{n}(x)|$ converges only pointwise, for all $x\in \mathbb{R}$?  I tried limiting the domain to $(0,1)$ and using Dirichlet Eta function, but that approach failed. I know that the sequence $f_{n}$ cannot be strictly positive and can't be equibounded, otherwise  $\sum \limits_{k=1}^{\infty}|f_{n}(x)|$ would converge uniformly by Weierstrass M-test.
120704
Find an example of a function sum $\sum_{n=1}^\infty u_n $ that converges   uniformly on $[a,b]$ and converges absolutely pointwise, but   $\sum_{n=1}^\infty |u_n| $ doesn't converge uniformly on $[a,b]$.   This seems pretty straightfoawrd but it's really hard to find an example. The choice of $u_n$ must be very specific. You can actually prove any power series wont work (due to Abel's theorem)  In addition, in order to prove the uniform convergence, $u_n$ must not satisfy Weierstrass' M-Test (because then $\sum |u_n|$ would converge uniformly aswell)
24565
Let $f:\mathbb{R}\to\mathbb{R}$ continuous such that $\lim_{x\to\infty}f(x+1)-f(x)=l$.  How can I prove that $\dfrac{f(x)}x$ converges for $x\to+\infty$ ?  I am just trying to prove the convergence, not that it converges to $l$ (that part be deduced immediately by Stolz-Cesaro and the sequential characterization of limits)
108976
Let $f_n(x):[0,1] \to \Bbb R$ be a sequence of differentiable functions such that for every $n$ and every $x \in [0,1]$ we have $|f'_n(x)| \leq 1$. Also, $f_n \to f$ (pointwise convergence) in $[0,1]$. Prove that $f_n$ converges to $f$ uniformly.   I managed to prove that for every $x_0$, there is a uniform convergence in $(x_0-\delta, x_0 +\delta)$ for some $\delta &gt; 0$.  How can I finish the proof?
106081
Does the series $\sum_{n=1}^{\infty}|x|^\sqrt n$ converge pointwise?
109844
How can one prove that $$f(z)=\sum_{n=1}^{\infty}\dfrac{z}{n(1+n|z|^{2})}$$ converges uniformly on $\mathbb{C}$ ?     My attempt was to let $z=x+iy$, then consider $$|\dfrac{z}{n(1+n|z|^{2})}|=\dfrac{\sqrt{x^{2}+y^{2}}}{\sqrt{(n^{2}x^{2}+n)^{2}+(n^{2}y)^{2}}}.$$     which looks like a mess. The main idea was to try to find an upper bound, $M_{k}$ so that I can apply the Weierstrass M-test. If I can show that the upper bound, $M_{k}$ is convergent (which is often simple by using root test etc.), I will be able to conclude that $f(z)$ converges uniformly.    Is there a simpler or more elegant way I can approach this question and if my approach is correct, can someone kindly guide me to find the required upper bound?
61698
This is a nice problem on series convergence that I recently stumbled upon. Given a non-negative sequence of real numbers $(a_n)$ such that $$\sum_{n=1}^\infty a_n &lt; \infty,$$ show that there exists a non-decreasing sequence of non-negative numbers $b_n$ such that $$b_n \to \infty \quad\text{ and } \quad \sum_{n=1}^\infty a_n b_n &lt; \infty.$$ In other words, for every convergent series with non-negative terms, there is another convergent series with "substantially larger" terms.  I have a solution (see below), but maybe someone else has a different, simpler, and/or more elegant solution.
55965
Here is an exercise, on analysis which i am stuck.    How do I prove that if $F_{n}(x)=\sum\limits_{k=1}^{n} \frac{\sin{kx}}{k}$, then the sequence $\{F_{n}(x)\}$ is boundedly convergent on $\mathbb{R}$?
104157
If $f(z)$ is analytic in the unit disk and $f(0)=0$, show that $$f(z)+f(z^2)+\cdots+ f(z^n)+\cdots$$ converges locally uniformly to an analytic function in the unit disk.  I am thinking to apply Weierstrass $M$ test for convergence or since all the $f(z^n)$ are analytic in the unit disk, so each of them can be expressed as the taylor series. Then rewriting the series again might work. But I am not sure. Any suggestions!
40818
I've encountered a question I couldn't answer, and I would appreciate any help: Is it true that $$f_n \xrightarrow{m}0\,\,\implies \,\,\frac{1}{n} \sum_{k=1}^{n}f_k \xrightarrow{m}0$$ where $(X,{\mathcal{B}},m)$ is a probability space, and $\xrightarrow{m}$ means: $\forall \epsilon&gt;0,\,  m\left(|f_n|\geq \epsilon\right) \rightarrow 0.$  This is similar to Cesaro convergence, which is weaker than regular convergence, but it is not the case here. I've tried to prove the proposition, or to give a counter example, but didn't manage to succeed in either. Any insights will be appreciated!
49344
Check that given series converges uniformly for $x\in(-\pi,\pi)$ or not  1) $\sum_{n=1}^{\infty} (\frac{x}{n})^n$  2) $\sum_{n=1}^{\infty}\frac{1}{((x+\pi)n)^2}$  first series is power series so I find the radius of convergence that comes $R=e$ which is inside the given set so I am not able to give the final conclusion. after this I were thinking to apply $M_n$ test , but I am not able to find $M_n$. Please help to solve this problem.
67083
Regarding this discussion here:   (thank you for the great answers, by the way)  I'm still trying to do Exercise 3.2.2 (b) from  by Dr. Pete Clark.  The question states that for the set $S = \mathbb{Z}^{+}$, the series $\sum_{i\in S}x_{i}$ converges unconditionally if and only if it converges absolutely, i.e, $\sum_{i\in S}|x_{i}| &lt; \infty$.  Thanks to the prior discussion, I get why unconditional convergence isn't simply the same thing as regular convergence in the case of $S = \mathbb{Z}^{+}$.  Unconditional convergence (to $x$) even allows for "cherry picking" of terms in the tail, rather than only being allowed to take them in order, and still reguires the resulting sums to stay within the $\epsilon$-band (of $x$).  (Again, thanks for the detailed answers to my previous question on this.)  My "best" attempt hits a brick wall:  Let $S = \mathbb{Z}^{+}$.  Write $S = S^{+}\cup S^{-}$, where $$S^{+} := \{n\in S : x_{n} \geq 0\}$$ and $$S^{-} := \{n\in S : x_{n} &lt; 0\}$$  $(\Rightarrow)$ Now if I assume that $\sum_{n=1}^{\infty}|x_{n}|$ is not finite ( that is, the series is not absolutely convergent ),  I want to show that the unconditional convergence definition fails.  I want to try to construct arbitrarily large finite subsets $K$ of $S^{+}$ or $S^{-}$ such that $\sum_{i\in K}x_{i}$ is arbitrarily large, which would prove the contrapositive.  I think I can do it if one of $S^{+}$ or $S^{-}$ is finite, but this need not be the case.  Any advice or suggestions?  I feel like I hit a brick wall.  Edit:  I added the solution as an answer below for anyone who might have this same question.  Thanks all for the help.
98761
This question arises because of a problem I was doing (Bartle 3rd edition, section 9.4 problem 3). It was like this. Given $a_n$ a decreasing sequence of positive numbers and suppose that $$\sum_{n=0}^{\infty}{a_n \sin{(nx)}}$$ Converge uniformly (It doesn't specify the domain, so I guess is for every x). Prove that $n a_n \to 0$. Clearly $\frac{1}{n}$ fits the description of $a_n$, and $n \frac{1}{n} \to 1 \neq 0$, so this would prove that there is a mistake in the problem if $\sum{\frac{\sin{(nx)}}{n}}$ converge uniformly for all $x$. So my question is if $\sum{\frac{\sin{(nx)}}{n}}$ converge uniformly for every $x$.  (I know that the series converge uniformly for every x in $[\delta, 2\pi - \delta]$, for $0 &lt; \delta &lt;2\pi$ by using the Dirichlet criterion.)
64951
Let $(a_n)_{n\in\mathbb{N}}$ be a sequence such that lim inf $|a_n|=0$. Prove that there is a subsequence $(a_{n_{k}})_{k\in\mathbb{N}}$ such that $\sum_{k=1}^\infty a_{n_k}$ converges.  I'm thought about showing that you can make a subsequence that is smaller then $1/n^2$. Is this the right strategy ? Any hints ? Should I use the Cauchy criterion ?
48575
I am wondering whether there exist a function $f(z) = \sum_{n\geq0} a_n z^n$ such that:   $f$ converges and is continuous on the closed unit disk $D$ and the series $\sum_n a_n z^n$ does not converge uniformly on $D$.   I have tried to construct a counter-example, but with no success so far.
52634
$\sum_0^\infty a_nz^n$ , $z\in \mathbb{C}$, a power series with $R:=\sup \{ t\ge0 : a_n t^n$ is bounded $\}$ as its convergence radius. I wish to prove that $\sum_0^\infty a_nz^n$ is locally normally convergent over $B(0,R)$.  What I did so far:   Let $r_1 &lt;r_2&lt;R$ , then for all $z\in B(0,r_1)$ :  $$\sum_0^\infty|a_nz^n| = \sum _0^\infty|a_n|\frac{r_2^n}{r_2^n}|z|^n \le \sum _0^\infty|a_n|r_2^n(\frac{r_1}{r_2})^n .$$  Now, because $r_2&lt;R$ , $|a_n|r_2^n &lt;M$ , so: $$\sum _0^\infty|a_n|r_2^n(\frac{r_1}{r_2})^n \le M\sum _0^\infty(\frac{r_1}{r_2})^n ,$$ which converges as a geometric series when $r_1&lt;r_2$. So any power series for $z\in B(0,r_1)$ is mutually bounded by $M\sum_0^\infty(\frac{r_1}{r_2})^n$.   Yet I don't succeed to make the next step, and show that  any $z\in B(0,r_1)$ has a neighborhood $U_z$ such that $\sum_0^\infty \sup_{U_z} |a_n z^n|$ converges.
13292
I must show that $\sum\limits_{n\geq1} \frac{\sin(nx)}{n^2}$ converges $\forall x \in {R}$. Then  if $f(x)=\sum\limits_{n=1}^\infty f_n(x)$, I must prove that  $f(x)$ is continuous for $x\in [0, \pi]$  and  $\int\limits_0^\pi f(x)=2\sum\limits_{n=1}^\infty \frac{1}{(2n-1)^3}$.  I used Weistrass M-test: $\sum\limits_{n\geq1} \frac{\sin(nx)}{n^2}$ converges uniformly because $|\frac{\sin(nx)}{n^2}|\le \frac{1}{n^2}, \forall x\in R$ and $\sum \frac {1}{n^2}$ converges.  I found somewhere that $f(x)=\lim\limits_{n\to\infty}f_n(x)$. Where does it come from? I don't understand. How $f(x)$ looks like? Without $f(x)$ I don't know how to prove the last equality from exercise:(    Later edit: If $ \sum\limits_{n\geq1}f_n\to f$ uniformly , then $f_n\to f$ (punctual converges to $f $- I hope this is correct translation) which means $f_n(x)\to f(x)$ when $n\to \infty ?$   I hope someone could help me again...
96575
I'm reading page number 4 here. In particular the section where it deals with the case $p=\infty$, that is , showing that $L^{\infty}$ is complete.    Two questions:  1) Why is the convergence uniform? where it says "for $x \in N^{c}$ , $f_{n}$ is a Cauchy sequence of complex numbers. Thus $f_{n} \rightarrow f$ uniformly. Clearly we have pointwise convergence but why is it uniform?  2) I don't see why $||f_{n} - f||_{\infty} \rightarrow 0$. Can you please explain this step?  Thanks.  (3/2015) Edit: The original link appears to be broken.  seems to provide a similar (maybe even identical) proof to the one the OP talks about with slight notational differences.