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In my Calculus II homework, I encountered the following exercise:

If the $n$th partial sum of a series $\sum_{n=1}^\infty a_n$ is $$s_n = \frac {n-1}{n+1}$$ find $a_n$ and $\sum_{n=1}^\infty a_n$.

I solved the exercise this way: ( I took $S_n=(n-1)/(n+1)$ to be equation (1))

enter image description here

Equations (2) and (3) answer the exercise's questions. However, when I tried to corroborate my answer in equation (3) by using the found $a_n$ in equation (2), I encountered an inconsistency that I haven't yet been able to harmonize. This is what I tried to do:

enter image description here

Equations (3) and (4) should have yielded the same answer but, this disparity I have been so far unable to harmonize. Your kind comments on what to do will be greatly appreciated.

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Your derivation of $a_n$ using $S_n - S_{n-1}$ is only valid for $n > 1$, i.e. $S_0$ is not defined, or rather, it is defined to be zero, but if you substitute zero in $S_n = \dfrac {n-1}{n+1}$ you get $-1$, which is the cause of the disparity.

We hence need to define $a_1$ separately: In fact $a_1 = S_1 = 0$. Hence:

$$S = \sum_{n=1}^\infty a_n = \sum_{n=\color{red}2}^\infty\frac 2{n(n+1)} = 1$$

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    $\begingroup$ General comment: in general an "empty sum" i.e. $\sum_{k=a}^b c_k$ where $b<a$ is defined to be zero. $\endgroup$ – Ian Feb 12 at 14:49
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This is not an answer, just a small addition to the post of player3236:

As already in the post of player3236 mentioned $s_n$ is only definied for all $n\in\mathbb{N}$. So you can't calculate $s_n-s_{n-1}$ for all $n\in\mathbb{N}$. But you can calculate $$s_{n+1}-s_n=a_{n+1}$$ without changing the range of n. By a similar calculation as you did above you will receive $a_{n+1}=\frac{2}{(n+1)(n+2)}$ for all $n\in\mathbb{N}$. Now you still have to find $a_1$, which can either be done directly by the given formular $a_1=s_1=\frac{1-1}{1+1}=0$ or by using the formular for $a_{n+1}$ $$s_2=\frac{1}{3}=\frac{2}{(1+1)(1+2)}=a_2 $$ and thus $a_1=0$ since $s_2=a_2+a_1$.

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