$(a_n)$ is bounded $\implies \sum a_n \frac{1}{n^2}$ converges

Question

Show that if $(a_n)$ is a bounded sequence, then

$$\sum \frac{1}{n^2} a_n $$ converges

I only could prove that this sequence is bounded, but not its convergence.

Thanks in advance.

Answer 1

Given that $|a_n|\le M$, then we can use the comparison test to show that because $$ \left|\,\frac{a_n}{n^2}\,\right|\le\frac M{n^2} $$ and that $\displaystyle\sum_{n=1}^\infty\frac1{n^2}$ converges, we also have that $\displaystyle\sum_{n=1}^\infty\frac{a_n}{n^2}$ converges absolutely.

Answer 2

In order to prove convergence it is sufficient to show that the sequence $\{S_M\}_{M\in\mathbb{N}^*}$ given by: $$ S_M = \sum_{n=1}^{M}\frac{a_n}{n^2}$$ is a Cauchy sequence. Since, assuming $M_1<M_2$: $$ S_{M_2}-S_{M_1} = \sum_{n=M_1+1}^{M_2}\frac{a_n}{n^2} $$ is bounded in absolute value by: $$ \sup_{n\in\mathbb{N}^*} |a_n|\cdot\sum_{n=M_1+1}^{M_2}\frac{1}{n^2}\leq \sup_{n\in\mathbb{N}^*} |a_n|\cdot\int_{M_1}^{M_2}\frac{dx}{x^2}=\sup_{n\in\mathbb{N}^*} |a_n|\cdot\left(\frac{1}{M_1}-\frac{1}{M_2}\right)$$ convergence follows.

Answer 3

$$V \le a_k \le U\tag{k > 0}$$ $$\frac{1}{k^2}V \le \frac{1}{k^2}a_k \le \frac{1}{k^2}U$$ $$\sum_{k=0}^\infty \frac{1}{k^2}V \le \sum_{k=0}^\infty \frac{1}{k^2}a_k \le \sum_{k=0}^\infty \frac{1}{k^2}U$$

$$\frac{\pi^2}{6}V \le \sum_{k=0}^\infty \frac{1}{k^2}a_k \le \frac{\pi^2}{6}U$$

Answer 4

One more way: Dirichlet test, which uses summation by parts. You have $0<a_n <M$ and $\frac{1}{n^2} \to_n 0$, so conditions are satisfied, and the series converges.

Answer 5

Use Dirichlet's test, which says: If the partial sum of the series summation (a_n) is bounded and the sequence (b_n) is a monotone sequence converging to zero, then the series summation (a_n b_n) is convergent.

Here (1/n^2) is a monotone sequence converging to zero.

Hint: 0< (1/n^2) < (1/n) and apply squeeze theorem to get the limit.