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How can one find the limit for the sequence $\{x_n\}^{+\infty}_{n=0}$ where $$x_0 = 0, x_1 = 1, x_{n+1} = \dfrac{x_n + nx_{n-1}}{n+1}$$ By computing the values I came to the conclusion that it converges to $\ln2$, but I can’t figure out how to solve this analytically.

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Hint $$x_{n+1}-x_{n}=-\dfrac{n}{n+1}(x_{n}-x_{n-1})\Longrightarrow (n+1)(x_{n+1}-x_{n})=-n(x_{n}-x_{n-1})$$ so $$(n+1)(x_{n+1}-x_{n})=(x_{1}-x_{0})(-1)^n=(-1)^n$$ so $$x_{n+1}-x_{n}=\dfrac{(-1)^n}{n+1}$$ so $$x_{n}=\sum_{i=1}^{n}(x_{i}-x_{i-1})+x_{0}=\sum_{i=1}^{n}\dfrac{(-1)^{i-1}}{i}$$ we konwn $$\ln{2}=1-\dfrac{1}{2}+\cdots+$$

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  • $\begingroup$ this answer are you ok? $\endgroup$ – math110 May 28 '15 at 13:53
  • $\begingroup$ Yes, that’s just right, many thanks! $\endgroup$ – macleginn May 28 '15 at 14:25
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Hint : Show $$x_n=\sum_{j=1}^n (-1)^{j+1}\frac{1}{j}$$

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  • $\begingroup$ I know that this the series expansion for ln2, but I failed to extract the difference relation as shown by math110. Thanks a lot, anyway! $\endgroup$ – macleginn May 28 '15 at 14:26

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