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. $a_n >0$

  • $\begingroup$ I think that $a_n>0$ should be a condition to make the question really interesting $\endgroup$
    – ajotatxe
    Jan 7, 2019 at 21:08
  • $\begingroup$ @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. $\endgroup$
    – amir na
    Jan 7, 2019 at 21:23
  • $\begingroup$ With the condition $a_n > 0$ it is a duplicate of math.stackexchange.com/q/1933001/42969. $\endgroup$
    – Martin R
    Jan 7, 2019 at 21:31
  • $\begingroup$ If $a_n>0$, by Cauchy-Schwarz $$\sum_{n=1}^{N}a_n\sum_{n=1}^{N}\frac{1}{n^2 a_n}\geq H_N^2\geq \log^2(N).$$ $\endgroup$ Jan 7, 2019 at 21:34

3 Answers 3


Perhaps a bit too trivial?



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.


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.


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