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

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. $$a_n >0$$

• I think that $a_n>0$ should be a condition to make the question really interesting – ajotatxe Jan 7 at 21:08
• @ajotatxe yes. The book I found the first question, said at the first of the series chapter, that for all questions, consider $a_n>0$ and I forgot to mention that. – amir na Jan 7 at 21:23
• With the condition $a_n > 0$ it is a duplicate of math.stackexchange.com/q/1933001/42969. – Martin R Jan 7 at 21:31
• If $a_n>0$, by Cauchy-Schwarz $$\sum_{n=1}^{N}a_n\sum_{n=1}^{N}\frac{1}{n^2 a_n}\geq H_N^2\geq \log^2(N).$$ – Jack D'Aurizio Jan 7 at 21:34

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