Find a sequence $\{a_n\}$ such that both $\sum {a_n}$ and $\sum{\frac{1}{n^2 a_n}}$ converge.

Question

Find a sequence $\{a_n\}$ such that both $\sum_1^\infty {a_n}$ and $\sum_1^\infty {\frac{1}{n^2 a_n}}$ converge. If no such sequence exists, prove that.

Actually the question was for $\sum_1^\infty {n{a_n}}$ and $\sum_1^\infty {\frac{1}{n^2 a_n}}$. For this problem; If we suppose that both are convergent, then their multiply $\sum_1^\infty 1/n$ must be convergent and this is a contradiction.

But for $\sum_1^\infty {a_n}$ and $\sum_1^\infty {\frac{1}{n^2 a_n}}$, if we multiply we get $\sum_1^\infty 1/n^2$ which is convergent. I tried to find a sequence such that both the series converge but I couldn't find any but also I couldn't prove that there is no such sequence exists.

Edit. $a_n >0$

Answer 1

The answer is yes, an example is $$a_n = \frac{(-1)^n}{n}$$ However, there is no such a sequence with positive terms. Indeed, if you suppsose that both $\sum a_n$ and $\sum\frac{1}{n^2a_n}$ are convergent and with positive terms, then you have by the AM-GM inequality $$\sum \frac{1}{n}=\sum \sqrt{a_n \cdot \frac{1}{n^2a_n}} \le \sum \frac{1}{2} \left(a_n + \frac{1}{n^2a_n} \right) < +\infty$$ which implies that the harmonic series is convergent: a contradiction.

Answer 2

Perhaps a bit too trivial?

$a_n=(-1)^n(1/n)$

Answer 3

The above answer shows that if the $a_n$s are allowed to change sign, then the sum may converge. However, if the $a_n$s are all positive, then by the Cauchy-Schwarz inequality,

\begin{align*} \sum_{n=1}^k\frac{1}{n}&=\sum_{n= 1}^k\frac{\sqrt{a_n}}{\sqrt{n^2a_n}}\\ &\leq \left(\sum_{n =1}^ka_n\right)^{1/2}\left(\sum_{n= 1}^k\frac{1}{n^2a_n}\right)^{1/2} \end{align*}

Since $\sum_{n=1}^k\frac{1}{n}\to\infty$ as $k\to\infty$, one of $\sum a_n$ and $\sum \frac{1}{n^2a_n}$ must diverge.