Series - The Basics — Question 2

Question 2 – Solution

Step 1: Form a finite partial sum.

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

Step 2: Identify the surviving boundary terms.

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

Step 3: Take the partial-sum limit.

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

Step 4: Compute and verify the remainder.

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.