# Find $\lim a_n$ if $a_{n+1}=a_n+\frac{a_n^2}{n^2}$ [duplicate]

$0\lt a_1\lt 1$, $a_{n+1}=a_n+\dfrac{a_n^2}{n^2}$,

Find $\lim\limits_{n\to\infty} a_n$

Well, $a_{n+1}\gt a_n$, But how to show that there is an upper bound of $\{a_n\}$ ? What is the limit ? Thanks a lot

## marked as duplicate by mickep, kingW3, Ivo Terek, Community♦Oct 4 '17 at 23:17

• @HirenGarai No, the limit is not $0$. For $a_1=1/2$ it seems to be $\ln 2$. Also, OP proved that the sequence is increasing... – mickep Oct 4 '17 at 14:32
• @mickep for $a_1 = 0.5$ limit seem to be $1.341...$ – Raghukul Raman Oct 4 '17 at 14:35
• does this sequence even converge?? I enumerated the value for $a_1 = 0.99999$ and I found that $a_{50000} = 35162$. – Raghukul Raman Oct 4 '17 at 14:43
• @RaghukulRaman, As shown in the link, if $a_1$ is close to $1$ then the limit is approximately $\frac{2}{1-a_1}$. In your case, this is $200000$ so your observation is not surprising. – Sangchul Lee Oct 4 '17 at 14:48