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If the nth partial Sum of the series $ \Sigma a_n$ is $s_n = \frac {n-1}{n+1}$ find $a_n$ and $\Sigma a_n$.

So from my understanding I need to use the vanishing condition which is a corollary of the div test (convergence/divergence tests) for series. I am really confused with this one, its a challenge question, but I just am not sure.

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  • $\begingroup$ A telescoping series will work fine. Notice $S_n = 1-\dfrac{2}{n+1}$. I'm sure you have encountered similar. $\endgroup$
    – David P
    Jul 23, 2017 at 21:52
  • $\begingroup$ @Rex: Hope this formatting is reflective of what you wanted. Roll it back if it is not. $\endgroup$
    – gary
    Jul 23, 2017 at 21:52
  • $\begingroup$ @gary, perfect formatting, thank you $\endgroup$
    – Rex
    Jul 23, 2017 at 21:55
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    $\begingroup$ Hint: $\,s_n=a_n+s_{n-1}\,$, so $a_n=s_n-s_{n-1}\,$. $\endgroup$
    – dxiv
    Jul 23, 2017 at 23:58

1 Answer 1

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$s_{n}=s_{n-1}+a_n\to a_n=s_n-s_{n-1}=\dfrac{n-1}{n+1}-\dfrac{n-1-1}{n-1+1}=\dfrac{n-1}{n+1}-\dfrac{n-2}{n}$

$a_n=\dfrac{2}{n^2+n}$

$s=\mathop {\lim }\limits_{n \to \infty } \, \dfrac{n-1}{n+1}=1$

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