Comparison and Limit Comparison Tests — Question 8

PDF ↗

Question 8

Consider ∑n=1∞1n+n\displaystyle\sum_{n=1}^{\infty}\frac1{n+\sqrt n}.

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

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

  3. Confirm the conclusion by limit comparison with 1/n1/n.

Original worksheet page 1: question and worked solution for 4-7-008
Show solutionHide solution

Question 8 – Solution

Step 1: Bound the denominator.

For n≥1n\ge1, n≤n\sqrt n\le n, so n+n≤2n.n+\sqrt n\le2n. Because both denominators are positive, taking reciprocals reverses the inequality: 1n+n≥12n.\frac1{n+\sqrt n}\ge\frac1{2n}.

Step 2: Apply Direct Comparison.

The smaller series ∑1/(2n)\sum1/(2n) diverges, so the larger given positive-term series diverges.

Step 3: Confirm by limit comparison.

limn→∞1/(n+n)1/n=limn→∞11+1/n=1.\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.

Original worksheet page 2: question and worked solution for 4-7-008

Original worksheet layout. Use Enlarge or open the PDF for a closer view.