Sequence of continuous functions, integral, series convergence Let $f_k$ be a sequence of continuous functions on $[0,1]$ such that $\int _0 ^1 f_k(x)x^ndx  = \int _0^1 x^{n+k} dx$ for all $n \in \mathbb{N}$.
Is $\sum _{k=1} ^{\infty}f_k(x)$ convergent?
Could you tell me how to solve this? I would appreciate all the hints.
Thank you.
 A: No. Hint: The right-hand-side $\int_0^1 x^{n+k} dx=c_k$ is a constant for each $k$. You can make $f_k(x)$ completely arbitrary on the interval $[0,9/10]$, and then adjust it on the interval $[9/10,1]$ to give the integral $\int_0^1 f_k(x)x^n dx$ the value $c_k$. Choose $f_k$ on the interval $[0,9/10]$ in such a way that the series diverges for any $0\leq x\leq 9/10$. If instead of always "adjusting" $f_k$ on the interval $[9/10,1]$, and you instead adjust it somewhere else for different $k$, you can construct a sequence $f_k$ which makes the series diverge for all $x\in[0,1]$.
EDIT I'll flesh out the hint. For the first case: Let $f_k(x)=1$ for $x\in[0,9/10]$, and let $f_k(x)$ on the interval $[0,9/10]$ be a straight line from the point $(9/10,1)$ to a point $(1,a_k)$. If $a_k$ were $1$, then $f_k$ would be identically 1 on the entire interval, and we would get $\int_0^1 f_k(x) x^n dx = \int_0^1 x^n dx > \int_0^1 x^{n+k} dx=c_k$. On the other hand, if we let $a_k\to-\infty$, then the integral $\int_0^1 f_k(x) x^n dx$ will tend to $-\infty$, which is less than $c_k>0$, so by the intermediate value theorem, there must exist some $a_k$ such that $\int_0^1 f_k(x) x^n dx$ is equal to $c_k$. Now the series $\sum_{k=1}^\infty f_k(x)$ will diverge for all $x\leq 9/10$ since it will just be the series $\sum 1=\infty$.
If we want the series to diverge for all $x$, then we could repeat the idea above, but instead of letting $f_k$ be $1$ on $[0,9/10]$, we could let it be $1$ on $[0,1-1/k]$ instead. Then the series would diverge for all $x<1$, since for each such $x$, all but finitely many terms of the series will be $1$. For $x=1$ one gets the series $\sum a_k$, and one needs to check that this series necessarily diverges. I will leave that as an exercise to you.
