How to prove if $\int^{\infty}_{0}f(x)dx$ a converges, then there is increasing sequence $x_n$, $\lim_{n \to \infty}f(x_n)=0$ I tried to prove it directly, but examples like $\sin(x^{2})$ makes it impossible to find the proper subsequence $x_{n}$;
I also tried proving by contraposition, but the converse negative statement is not very clear and have too many cases.
 A: In this answer, it is assumed that $f$ is continuous, because otherwise a counter-example can be constructed as Did noticed.
The sequence  $ (z_n)_{n\ge0}$ with $z_n=\int_{n}^{n+1} f(x)dx$ converges to zero, because the integral is convergent. Moreover, by the intermediat value theorem, for each $n$ there exists an $ x_n \in [n,n+1]$ such that
$z_n=f(x_n)$. The sequence   $ (x_n)_{n\ge0}$ satisfies the desired requirements.
A: Just to show why continuity of $f$ is necessary: For $n\in N$ let $A(n)$ be the largest $m\in N$ such that $m> n$ and $\sum_{j= n}^m (1/j)<1.$ For $n,k\in N$ with $k\leq A(n),$ let $S(n,k)=n-1+\sum_{j=0}^k(1/(n+j)$ and let $S(n,0)=n-1.$
For $n,k\in N$ with $k\leq A(n)$ let $f(x)=(-1)^{n+k}$  when $x\in [S(n,k-1),S(n,k). $ 
For $x\in [S(n,A_n),n)$ let $f(x)=(-1)^{n+A(n)+1}.$ 
We can  show that $(I_n)_{n\in N}=(\int_{n-1}^nf(x)\;dx)_{n\in N}$ is an alternating series and that $(|I_n|)_{n\in N}$ is monotonically converging to $0.$ And for $n\in N$ and $y\in [n,n+1)$ we have $\int_0^yf(x)\;dx=(\sum_{j=0}^n I_j)+O(1/n).$
