# Why does this Infinite Series have contradictory convergences?

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))

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:

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.

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:

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

• General comment: in general an "empty sum" i.e. $\sum_{k=a}^b c_k$ where $b<a$ is defined to be zero.
– Ian
Commented Feb 12, 2021 at 14:49

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 $$s_2=a_2+a_1$$.