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

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

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

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    $\begingroup$ @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... $\endgroup$ – mickep Oct 4 '17 at 14:32
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    $\begingroup$ There are a couple of postings which answer the convergence part, see this and this. Regarding the closed-form part, I once studied the limit as function of the initial term, and if I remember correctly the function seems to develop a lacunary boundary. This makes me suspect that no simple closed form can ever be found. $\endgroup$ – Sangchul Lee Oct 4 '17 at 14:35
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    $\begingroup$ @mickep for $a_1 = 0.5 $ limit seem to be $1.341...$ $\endgroup$ – Raghukul Raman Oct 4 '17 at 14:35
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    $\begingroup$ does this sequence even converge?? I enumerated the value for $a_1 = 0.99999$ and I found that $a_{50000} = 35162$. $\endgroup$ – Raghukul Raman Oct 4 '17 at 14:43
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    $\begingroup$ @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. $\endgroup$ – Sangchul Lee Oct 4 '17 at 14:48