Convergence and Divergence of Series — Question 1

Question 1 – Solution

Step 1: Check the necessary condition.

Let $a_n=1/\sqrt n$. Since $\sqrt n\to\infty$, $a_n\to0$. This necessary condition is satisfied, so the nth-term test is inconclusive; it can prove divergence when the limit is nonzero, but it can never prove convergence.

Step 2: Choose a grouping strategy.

The terms decrease too slowly for a geometric comparison. Grouping between consecutive squares makes their sizes and the number of terms in each block easy to control. For $k\ge1$, consider $k^2\le n<(k+1)^2$. There are $$(k+1)^2-k^2=2k+1$$ terms, and every one satisfies $\sqrt n<k+1$, hence $1/\sqrt n>1/(k+1)$. Therefore the $k$th block has sum greater than $$\frac{2k+1}{k+1}>1.$$

Step 3: Convert the block estimate into divergence.

The partial sum through $(m+1)^2-1$ contains $m$ such blocks and consequently exceeds $m$. Since $m$ can be arbitrarily large, the increasing sequence of partial sums is unbounded.

Step 4: State and check the conclusion.

Therefore $$\boxed{\displaystyle\sum_{n=1}^{\infty}\frac1{\sqrt n}\text{ diverges}.}$$ This agrees with the $p$-series rule: $p=1/2\le1$.