Let $\left \{ a_n \right \}$ be a positive sequence that converges to zero. And let $g$ be an integrable function, continuous at zero such that $g(0)\ne0$.

Prove: $\sum_{n=1}^{\infty}a_n$ converges iff $\sum_{n=1}^\infty\int_0^{a_n}g(t)\,dt$ converges.

My attempt: I try to use the integral test but I don't know how to continue from there and moreover how to use that $g(0)\ne0$ and $g$ is continuous at zero. Any hints would be appreciated.


1 Answer 1


Assume $\sum a_n $ converges. Then, using the continuity of $g$ at 0 and the fact that $a_n \to 0$, we have that for $n \in \mathbb{N}$ sufficiently large, $$ (g(0) - 1)a_n \leq \int_0^{a_n} g (t) dt \leq (g(0) + 1)a_n. $$
Since both sides of the inequality converge when you sum over $n$, we get the convergence of the series with integrals.

Now the opposite, assume $\sum \int_0^{a_n} g (t) dt$ converges , and without loss of generality assume $g(0) > 0$. Then, from the continuity of $g$ at 0, and the fact that $a_n \to 0$, we have $\int_0^{a_n} g(t) dt \geq ( g(0) - \epsilon ) a_n $ for $n$ large enough, where $\epsilon >0$ is fixed so that $( g(0) - \epsilon ) >0$. Now the series with integrals, which converges, dominates the series with $a_n$, hence the convergence of $\sum a_n$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .