Does the series $$\sum_{n=1}^{\infty}\frac{a_n}{a_{n+1}}\frac{1}{n}$$ diverge for any $a_n$, satisfying $0<a_n<1$, $n=1, 2, 3\dots$ ?

  • 1
    $\begingroup$ It is an interesting problem. Do you know the source? $\endgroup$ Jun 1, 2014 at 15:28
  • $\begingroup$ I came to this problem from another problem of Monthly Journal. $\endgroup$
    – Mher
    Jun 2, 2014 at 12:07
  • $\begingroup$ Maybe you could say it in the body of the question. $\endgroup$ Jun 2, 2014 at 19:11

1 Answer 1


For each $n$, let $b_n\in (-\infty,0)$ such that $\exp(b_n)=a_n$. By convexity of the exponential function on $\mathbb R$, $$\tag{1}\exp\left(\frac 1M\sum_{j=1}^M(b_j-b_{j+1})\right)\leqslant \frac 1M\sum_{j=1}^M\exp(b_j-b_{j+1})=\frac 1M\sum_{j=1}^M\frac{a_j}{a_{j+1}}.$$ Assume that $\sum_{j=1}^\infty\frac{a_j}{a_{j+1}}\frac 1j$ is convergent. Since for each $N\leqslant M$, $$\frac 1M\sum_{j=1}^M\frac{a_j}{a_{j+1}}\leqslant \frac 1M\sum_{j=1}^N\frac{a_j}{a_{j+1}}+\sum_{j=N}^M\frac{a_j}{a_{j+1}}\underbrace{\frac 1M}_{\leqslant \frac 1j}\leqslant\frac 1M\sum_{j=1}^N\frac{a_j}{a_{j+1}}+\sum_{j=N}^M\frac{a_j}{a_{j+1}}\frac 1j,$$ we obtain that $$\limsup_{M\to +\infty}\frac 1M\sum_{j=1}^M\frac{a_j}{a_{j+1}}\leqslant\sum_{j=N}^\infty\frac{a_j}{a_{j+1}}\frac 1j,$$ hence by (1), $$\lim_{M\to +\infty}\exp\left(\frac 1M\sum_{j=1}^M(b_j-b_{j+1})\right)=0.$$ Since $\sum_{j=1}^M(b_j-b_{j+1})=b_1-b_{M+1}$, it follows that $$\lim_{M\to +\infty}\exp(-b_{M+1}/M)=0$$ which is not possible because each $b_M$ is negative.

The argument would boil down if we had considered $\sum_{n=1}^\infty\frac{a_{n+1}}{na_n}$ and in this case the series may converge, for example with $a_n:=2^{-n^2}$.


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