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Let $a_0=\pi$, $a_1=\pi^2$, $a_{n+1}=a_n+\frac{2a_{n-1}}{n+1}$ How can we prove that $a_n/n^2$ converges?

It is easy to see that $$a_n-a_1=\sum_{k=1}^{n-1}\frac{2a_{k-1}}{k+1}\geq 2a_0\sum_{k=1}^{n-1}\frac{1}{k+1}\to\infty.$$ Then how to do?

xldd
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  • @Did: May I ask what is the point of closing a question, by marking it as a duplicate of a closed question? Cui prodest? – Jack D'Aurizio Nov 01 '16 at 19:13
  • @JackD'Aurizio The point of closing this question, by marking it as a duplicate of a closed question, is that it is a duplicate of the other question. Why do you ask? (Interestingly, your comment was not pinged to me. Might be due to the : after the user name.) – Did Nov 02 '16 at 15:44
  • @Did : I am glad Mr.Obvious gave his contribute, but my hidden point was: if we close a question by marking it as a duplicate of a closed question, we leave no active "reference" for new askers. Are we sure it is a good practice to follow? – Jack D'Aurizio Nov 02 '16 at 15:46
  • @JackD'Aurizio Please calm down. Anyway, I can see that you are actively engaged in reopening the other question so what is the point of this discussion? (Apparently, a space before : re-activates the ping.) – Did Nov 02 '16 at 15:49
  • @Did : I am absolutely calm, and sorry if I gave you the impression of over-heating. I just wanted to the discuss about such strange situations (closing by marking as a duplicate of a closed question) because my personal opinion is that we should care about leaving active references, too. I pinged you since I saw you voted for closing, just that. – Jack D'Aurizio Nov 02 '16 at 15:53
  • @JackD'Aurizio OK. Apparently the system does allow to elect a closed question as duplicate, dunno if this is on purpose but it does. – Did Nov 02 '16 at 15:55

1 Answers1

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Let $(b_n)=(\frac {a_n} {n^2})$, we have that \begin{align} b_{n+1}-b_n&=\frac {n^2a_{n+1}-(n+1)^2a_n}{n^2(n+1)^2}\\ &=\frac {n^2(a_{n+1}-a_n)-(2n+1)a_n}{n^2(n+1)^2}\\ &=\frac {\frac {2n^2}{n+1}a_{n-1}-(2n+1)a_n}{n^2(n+1)^2}\\ &=\frac {2n^2a_{n-1}-(2n+1)(n+1)a_n}{n^2(n+1)^3}\\ &=\frac {2n^2(a_{n-1}-a_n)-(3n+1)a_n}{n^2(n+1)^3}\\ &\le0 \end{align} Since $(b_n)\geq0$ and $(b_n)$ is decreasing, $(b_n)$ is convergent.

Aforest
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