2

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

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

2016
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  • It is a sequence of positive numbers.So if we assume that the limit exits , and calculate the limit of $\frac{a_{n+1}}{a_n}$ , then it is easily seen to be 0 , and so the required limit is 0. – hiren_garai Oct 04 '17 at 14:30
  • @HirenGarai How does one arrive at that conclusion? It isn't correct. – Mark Viola Oct 04 '17 at 14:32
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    @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 04 '17 at 14:32
  • Yes, My mistake, the limit of $\frac{a_{n+1}}{a_n}$ is not 0, .. so the required limit is not 0.. – hiren_garai Oct 04 '17 at 14:34
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    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. – Sangchul Lee Oct 04 '17 at 14:35
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    @mickep for $a_1 = 0.5 $ limit seem to be $1.341...$ – Raghukul Raman Oct 04 '17 at 14:35
  • @RaghukulRaman Indeed, my value was wrong. I accidentaly looked at $a_n=a_{n-1}+a_{n-1}^2/n^2$ or something similar. Well, then I take that part of my comment back. Thanks! – mickep Oct 04 '17 at 14:39
  • @SangchulLee Thank you very much! I searched, but no results....... – 2016 Oct 04 '17 at 14:39
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    does this sequence even converge?? I enumerated the value for $a_1 = 0.99999$ and I found that $a_{50000} = 35162$. – Raghukul Raman Oct 04 '17 at 14:43
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    @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 04 '17 at 14:48

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