Comparison and Limit Comparison Tests — Question 8

*Question 8 Consider $\displaystyle\sum_{n=1}^{\infty}\frac1{n+\sqrt n}$.

1. Prove a direct lower bound using $\sqrt n\le n$.

2. Use the correct comparison direction to classify the series.

3. Confirm the conclusion by limit comparison with $1/n$.

*Question 8 – Solution

For $n\ge1$, $\sqrt n\le n$, so $$n+\sqrt n\le2n.$$ Because both denominators are positive, taking reciprocals reverses the inequality: $$\frac1{n+\sqrt n}\ge\frac1{2n}.$$

The smaller series $\sum1/(2n)$ diverges, so the larger given positive-term series diverges.

$$\lim_{n\to\infty}\frac{1/(n+\sqrt n)}{1/n} =\lim_{n\to\infty}\frac1{1+1/\sqrt n}=1.$$ This finite positive limit confirms harmonic tail behavior.