Series - The Basics — Question 2

*Question 2 Consider $\displaystyle\sum_{n=1}^{\infty}\left(\frac1n-\frac1{n+1}\right)$.

1. Write out the first four terms and display the cancellation.

2. Find a formula for the $N$th partial sum, identifying the surviving boundary terms.

3. Decide whether the series converges and, if so, find its sum and exact remainder.

*Question 2 – Solution

Never telescope directly to infinity. First write the $N$th partial sum: $$\begin{aligned} s_N&=\left(1-\frac12\right)+\left(\frac12-\frac13\right) +\left(\frac13-\frac14\right)+\cdots+\left(\frac1N-\frac1{N+1}\right)\\ &=1-\frac1{N+1}. \end{aligned}$$

Every interior fraction appears once positively and once negatively, so it cancels. Only the initial $1$ and final $-1/(N+1)$ survive.

By definition, the series converges exactly when $(s_N)$ has a finite limit. Here $$S=\lim_{N\to\infty}s_N=1.$$

The exact tail after $N$ terms is $$R_N=S-s_N=\frac1{N+1}.$$ As an algebraic check, $s_N-s_{N-1}=1/N-1/(N+1)$, which reproduces the original $N$th summand exactly.