Series - The Basics — Question 9

*Question 9 Consider $\displaystyle\sum_{n=1}^{\infty}(\sqrt{n+1}-\sqrt n)$.

1. Write the first four terms and expose the telescoping cancellation.

2. Derive the $N$th partial sum and determine whether the series converges.

3. Reconcile the divergence with the fact that the individual terms approach zero.

*Question 9 – Solution

Write the $N$th partial sum before passing to infinity: $$\begin{aligned} s_N&=(\sqrt2-1)+(\sqrt3-\sqrt2)+\cdots+(\sqrt{N+1}-\sqrt N)\\ &=\sqrt{N+1}-1. \end{aligned}$$

All interior square-root terms cancel, leaving only the first negative boundary term and final positive boundary term. Because $\sqrt{N+1}-1\to\infty$, the partial sums are unbounded and the series diverges to $+\infty$.

The summand does approach zero. Rationalization gives $$\sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n}\longrightarrow0.$$

There is no contradiction: the nth-term condition is necessary, not sufficient. Here every increment is positive and shrinks to zero, but the accumulated total still grows without bound. Since the denominator above is asymptotic to $2\sqrt n$, each increment is asymptotic to $1/(2\sqrt n)$, consistent with a divergent $p$-series scale.