If series is not uniformly convergent, can we still integrate term by term? We know that if $\sum a_nx^n$ converges uniformly, then we can integrate term by term. So this is just a sufficient condition, right? Does there exists a series not converging uniformly and still we can perform term by term integration? I am looking for some general results in this direction.
 A: With a power series like that, there's a radius of convergence $r$ - and for any $s<r$, the series converges uniformly for $|x|<s$. With that, we can use the uniform convergence to integrate term by term with the same radius of convergence.
The only place we can't use uniform convergence? The boundary.
And on that boundary... Abel's theorem. If the series converges at a point on the boundary, then it's the limit of the values inside, approaching head-on. For example, this theorem applied to the series $\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$ gives us the alternating harmonic series $\ln(2)=1-\frac12+\frac13-\frac14+\cdots$.
Relating to the term-by-term integration your interested in, that logarithm series is the integral of $\frac1{1+x}=1-x+x^2-x^3+\cdots$, and we then have
$$\int_0^1 1-x+x^2-x^3+\cdots\,dx = \left.x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots\right|_{x=1} = \ln(2)$$
A: In response to a comment by the proposer to another A:
For $n\in \Bbb Z^+$ let $g_n(x)=0$ for $x\not \in [2^{-n-1},2^{-n}]$ and let $g_n$ be integrable on $[2^{-n-1},2^{-n}]$ such that $\sup \{|g_n(x)|: x\in [2^{-n-1},2^{-n}]\}\ge 1.$ And such that $\lim_{n\to \infty}\int_0^1g_n(t)dt=0.$
Let $g_0(x)=0$ for all $x.$
For $n\ge 0$ let $f_n=g_n-g_{n+1} .$
Then $\sum_{j=0}^nf_j=g_0-g_n=-g_n$ converges point-wise, but not uniformly, to $0$. And $\sum_{j=0}^n\int_0^1f_j(t)dt=-\int_0^1 g_n(t)dt$ converges to $0.$
For example, for $n\in \Bbb Z^+:$ Let $a_n=(2^{-n-1}+2^{-n})/2.$ Let $g_n$ be linear on $[2^{-n-1},a_n]$ and on $[a_n,2^{-n}]$ with $g_n(2^{-n-1})=0=g_n(2^{-n})$ and $g_n(a_n)=n.$
You can easily produce variations on this theme.
