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If the $n^{th}$ partial sum of a series $\sum a_n$ is $S_n = \dfrac{n-1}{n+1}$, find $a_n$ and the sum.

By definition, the sum of the series is the $\lim n\to\infty$ of it's $n^{th}$ partial sum.

$$\text{ Sum = } \lim_{n\to\infty} = \dfrac{n-1}{n+1} = 1$$

I am asked to find $a_n$. How do I do this? What is the procedure?

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    $\begingroup$ Hint:$$S_n-S_{n-1}=~?$$(by the definition of a partial sum of course) $\endgroup$ – Simply Beautiful Art Mar 22 '17 at 22:20
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    $\begingroup$ I believe it is $a_n$! :o $\endgroup$ – K Split X Mar 22 '17 at 22:22
  • $\begingroup$ :-) Well there you go! $\endgroup$ – Simply Beautiful Art Mar 22 '17 at 22:23
  • $\begingroup$ Really nice hint thanks! $\endgroup$ – K Split X Mar 22 '17 at 22:24
  • $\begingroup$ @SimplyBeautifulArt There is the issue of a seed value, else the difference equation is not fully specified. ;-)) $\endgroup$ – Mark Viola Mar 22 '17 at 23:00
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Let $S_n$ be the partial sum of a sequence $a_n$. Then, we can write

$$S_n=\sum_{k=1}^na_k \tag1$$

for $n\ge 1$.

Next, using the hint in the comment from S.B. Art, we have from $(1)$

$$a_n = S_{n}-S_{n-1}=\frac{2}{n}-\frac2{n+1} \tag2$$

for $n\ge 2$. The expression in $(2)$ is not valid for $n=1$ since $S_0$ is not defined by $(1)$.

However, we are given $S_n=\frac{n-1}{n+1}$. Thus, $S_1=0$ and hence $a_1=0$ also.

Finally, we can write

$$a_k=\begin{cases}\frac2k-\frac2{k+1}&,k\ge 2\\\\0&,k=1\end{cases}$$

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  • $\begingroup$ I don't understand how you got the $k = 1 \to 0$ part... using the $a_n$ that we found, why can't we just sub in $n=1$, which yields $$\dfrac{2}{1} - \dfrac{2}{1+1} = 2-1 =1$$ Why do you have it as another case? $\endgroup$ – K Split X Mar 22 '17 at 23:07
  • $\begingroup$ Is it because we use $S_{n} - S_{n-1}$, but we get $S_{1} - S_{0}$, and $S_{0}$ is not defined. Is that the reasoning? $\endgroup$ – K Split X Mar 22 '17 at 23:09
  • $\begingroup$ @KSplitX Yes, that is precisely the reason. Note that if we simply use $a_k=\frac1k-\frac1{k+1}$ for all $k\ge 1$, then $S_n=2-\frac{2}{n+1}$. $\endgroup$ – Mark Viola Mar 22 '17 at 23:13

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