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If the $n^{th}$ partial sum of a series $\sum a_n$ is $S_n = \frac{n−1}{n+1}$ then find $a_n$ and find the sum of this series. Fully justify your answer.

Using the definition.

$a_n = S_{n} - S_{n-1} $

$$a_n = \frac{n-1}{n+1} - \frac{n-2}{n}$$

$$= \frac{n(n-1) - (n-2)(n+1)}{n(n+1)}$$

$$= \frac{n^2-n - (n^2-n-2)}{n(n+1)}$$

$$= \frac{2}{n(n+1)}$$

Hmm, I can use PFD on this to make it.

$$= \frac{2}{n} - \frac{2}{n+1}$$

Is this right? What about n = 0

The sum is 1 right? cus lim at $\infty$ ?

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marked as duplicate by dxiv, Claude Leibovici, polfosol, user91500, Shailesh Mar 28 '17 at 10:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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As you have shown, $a_i= \frac{2}{i}-\frac{2}{i+1}$ holds for $i\geq 2 $ and $a_1=S_1=0$. Now write, $$S_n= \sum_{i=1}^{n}a_i=\sum_{i=2}^n (\frac{2}{i}-\frac{2}{i+1})= (\frac{2}{2}-\frac{2}{n+1})= (1-\frac{2}{n+1})$$ Now calculate the limit to find the sum of the series. You don't need to worry about $i=0$ since $a_i$ is defined for $i=1,2,3 \cdots$

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  • $\begingroup$ The sum is one right? I was told that the limit at infinity of the partial sum, is the sum? $\endgroup$ – user349557 Mar 28 '17 at 6:26
  • $\begingroup$ @user349557, Sorry I made an error, I have corrected it now. Yes the limit is 1. It is clear now. $\endgroup$ – Parish Mar 28 '17 at 6:35
  • $\begingroup$ $a_0$ can be defined. $\endgroup$ – Yves Daoust Mar 28 '17 at 6:58
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Obviously, the sequence can be defined as

$$\begin{cases}a_0=S_0=-1,\\a_n=S_n-S_{n-1}=\dfrac{n-1}{n+1}-\dfrac{n-2}{n}=\dfrac2{n(n+1)},n>0.\end{cases}$$

(In particular, $a_1=1$, ensuring $S_1=a_0+a_1=0$.)

The sum is indeed $\lim_{n\to\infty}\dfrac{n+1-2}{n+1}=1$.

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  • $\begingroup$ Whether the sequence is considered to start at $0$ or $1$ is a matter of taste, unless your course mandates one of the two conventions. But if you make it start at $i=1$, then $a_1=0$. Actually, the question doesn't have a single answer. $\endgroup$ – Yves Daoust Mar 28 '17 at 7:02
  • $\begingroup$ I agree absolutely, it is a matter of taste. There is no fundamental difference. $\endgroup$ – Parish Mar 28 '17 at 7:07
  • $\begingroup$ @Parish: the difference is $a_1=0$ vs. $a_1=1$, which is not so innocuous. $\endgroup$ – Yves Daoust Mar 28 '17 at 7:09

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