Does $\sum f_n$ converge uniformly on an interval $[a, b]$

Suppose $f_n$ are continuous real valued functions on $[a, b]$ such that $\sum_{n=1}^\infty f_n$ converges; does it follow that the sequence $a_N =\sum_{n=1}^N f_n$ converges uniformly?

Since $f_n$ is bounded on the compact set $[a,b]$, does uniform convergence follow by the Weierstrass M-test?

• you should enclose your thoughts, attempts, work. You could also search MSE, similar questions have already been answered. Also, think of an example, $f_n(x)=x^n$. Is $f$ required to be continuous? – Mirko Dec 12 '15 at 5:58
• I think putting this on hold is a bit harsh. OP clearly stated that he thinks the series is uniformly convergent by the Weierstrass M-test. He missed a hypothesis of the test, but that doesn't mean OP didn't think about the problem. – Seven Dec 12 '15 at 10:15
• @Seven Notice that the question was edited. Before the edit, there was no mention of M test. – user223391 Dec 12 '15 at 16:47
• @avid19 Oh. I saw the question only after that edit was made. – Seven Dec 12 '15 at 17:46

No, $\sum_{n=1}^\infty f_n$ may not converge uniformly. For example, take $a= 0$, $b= 1$ and

$f_1 = x$

$f_n = x^n - x^{n-1}$ for $n \geq 2$.

Then $\sum_{n=1}^N f_n = x^N$, and therefore $\sum_{n=1}^\infty f_n$ converges point-wise, but it does not converge uniformly on $[0,1]$.

Also, it is true that each $f_n$ is bounded, being continuous on a compact set. The problem with your argument about using the Weierstrass M-test is that the series of bounds on $f_n$ do not form a convergent series.

You can check that each $|f_n| \leq \frac{(n-1)^{(n-1)}}{n^n} = M_n$ (say), for $n \geq 2$, and you can't find a better bound. But $\sum_{n=1}^\infty M_n$ does not converge.

Hence, the Weierstrass M-test cannot be used here.