# Given a decreasing function s.t. $\int_0^\infty f(x)\,dx<\infty,$ prove $\sum_{n=1}^\infty f(na)$ converges

Let $f\in C([0,\infty))$ be a decreasing function such that $\int_0^\infty f(x)\,dx$ converges.

Prove $\sum_{n=1}^\infty f(na)$ converges, $\forall a>0$

My attempt:

By the Cauchy criterion, there exists $M>0,$ such that for $t-1>M:$

$$f(t)=\int_{t-1}^t f(t) \, dx \leq \int_{t-1}^t f(x)\,dx\xrightarrow{t \to \infty} 0$$

Hence, $f$ is non-negative.

$f$ is decreasing $\implies f(na)\leq f(a), \forall a>0, n\in \mathbb{N}.$

By integral monotonicity and non-negativity of $f$:

$$\int_1^\infty f(nx)\,dx \leq \int_1^\infty f(x)\,dx \leq \int_0^\infty f(x)\,dx$$

Hence $\int_1^\infty f(nx)\,dx$ converges and therefore $\sum_{n=1}^\infty f(na)$ converges.

Is that correct? If so, why is continuity necessary ? Is there a simpler way to prove it?

Any help appreciated.

• $$\int_{(n-1)a}^{na}f(x)dx\geq af(na)$$ – MAN-MADE Aug 22 '17 at 18:33

Hint: The following thing you wrote is the key:

$$f(t)=\int_{t-1}^{t}f(t)dx \leq \int_{t-1}^{t}f(x)dx$$

After that, just pick $t=an$ and sum over all $n$.

PS: and no, the assumption on continuity is not needed. Just integrability for the well-definedness.

As I commented before:

Since $f$ is decreasing and continuous, $$\int_{(n-1)a}^{na}f(x)dx\geq [na-(n-1)a]\inf_{(n-1)a< x\leq na}\{f(x)\}=af(na)$$

Then $$\int_{0}^{\infty}f(x)dx=\sum_{n=1}^{\infty}\int_{(n-1)a}^{na}f(x)dx\geq a\sum_{n=1}^{\infty}f(na)$$

Hence, since $a\in\mathbb{R}^+,$ $\sum_{n=1}^{\infty}f(na)<\infty$

\begin{align} & af(a) + af(2a) + af(3a) +\cdots \\[10pt] \le {} & \int_0^a f + \int_a^{2a} f + \int_{2a}^{3a} f + \cdots <\infty. \end{align}

This works if $f \ge0$ everywhere. If $f<0$ somewhere, then an easy argument shows the integral does not converge.

This is the integral test for convergence of series. Do a change of variable x --- to --- ax. The integral converges so does series g(n)=f(na).