Convergence and Divergence of Series — Question 6

Question 6 – Solution

Step 1: Check the necessary condition.

Let $a_n=e^{-\sqrt n}$. Because $\sqrt n\to\infty$, the exponent tends to $-\infty$ and $a_n\to0$. The nth-term condition holds but does not prove convergence.

Step 2: Organize the tail into square blocks.

In the block $k^2\le n<(k+1)^2$, there are exactly $2k+1$ indices. Also $\sqrt n\ge k$, so the decreasing exponential satisfies $e^{-\sqrt n}\le e^{-k}$. Therefore $$\sum_{n=k^2}^{(k+1)^2-1}e^{-\sqrt n}\le(2k+1)e^{-k}.$$

Step 3: Classify the series of block bounds.

It remains to check $\sum_{k=1}^{\infty}(2k+1)e^{-k}$. Let $q=e^{-1}<1$. Then $$2\sum_{k=1}^{\infty}kq^k+\sum_{k=1}^{\infty}q^k,$$ Both series converge: $\sum q^k$ is geometric, and $\sum kq^k=q/(1-q)^2$ follows by differentiating the geometric series.

Step 4: Conclude by comparison.

The original nonnegative series is bounded above block by block by a convergent series, so its increasing partial sums are bounded and $$\boxed{\displaystyle\sum_{n=1}^{\infty}e^{-\sqrt n}\text{ converges}.}$$