## Spring 2017, problem 39

Let $a_{n+1} = a_n + \frac{1}{{a_n}^{2017}}$ and $a_1=1$. Show that the sum $\sum^{\infty}_{n=1}{\frac{1}{n a_n}}$ converges.

### Comments

Hello Hubert !

Your method is well suited to the sequence (an) defined in the problem .Suppose now that we define another sequence (bn) which diverges too very slowly. It seems to me that the sum of 1/n(bn) will also converge but, if this is true,what sort of proof would you use ? More generally is the sum of 1/nf(n) from 1 to infinity converging with a function f slowly growing to infinity ?

In fact I was too quick to write this comment because after some more thoughts I found out that even if the sequence (bn) diverges slowly, the sum of 1/nb(n) might not converge. If, for example, the sequence behaves like Log(n) when n tends to infinity then the sum of 1/nLog(n) diverges like Log(Log(n)). At the same time a behaviour like Log(n)^2 is enough to make the sum converge ! In other words there are no general rules.

First I'm so happy to see some analysis here, my friends,,

now $(a_n)$ increases with $(a_n\rightarrow l\in\mathbb{R}\Rightarrow l=l+\frac 1 {l^\alpha})\Rightarrow l=+\infty$

$\forall \beta, \; (1+\epsilon)^\beta-1\stackrel 0 \sim \beta \epsilon\Rightarrow~~b_n=a_{n+1}^\beta-a_n^\beta\; \stackrel \infty \sim \; \beta {a_n^{\beta-\alpha-1}}, \; \alpha=2017$

now a kind of "telescopic Cesaro" yields $\beta=\alpha+1\Rightarrow~~a_n^\beta\sim\sum b_n\sim\beta n\Rightarrow~~$

$a_n\sim(\beta n)^{\frac 1 \beta}\Rightarrow~~\frac 1 {na_n}\sim \frac \gamma {n^{1+\frac 1 \beta}}, \; \gamma>0$, which converges by $1+\frac 1 \beta>1$.