Problem 9 - Analysis with an introduction to proof

Question

Determine whether or not the series \(\sum_{n=1}^{\infty} 1 /(\sqrt{n+1}+\sqrt{n})\) converges. Justify your answer. ??

Step-by-Step Solution

Answer

The given series \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n+1}+\sqrt{n}}\) diverges. This is determined by applying the Comparison Test and finding a simpler comparison series \(\sum_{n=1}^{\infty} \frac{1}{2\sqrt{n}}\) which diverges. Since the terms of the comparison series are greater than or equal to the terms of the given series, the original series also diverges.

Step 1: Write down the series

The given series is: \[\sum_{n=1}^{\infty} \frac{1}{\sqrt{n+1}+\sqrt{n}}\]

Step 2: Apply the Comparison Test

To apply the Comparison Test, we need to find a simpler series that has terms greater than or equal to the terms of the given series, and whose convergence can be easily determined. Note that: \[\sqrt{n+1}+\sqrt{n} > \sqrt{n}+\sqrt{n} = 2\sqrt{n}\] Thus, we have: \[\frac{1}{\sqrt{n+1}+\sqrt{n}} < \frac{1}{2\sqrt{n}}\] Now consider the series: \[\sum_{n=1}^{\infty} \frac{1}{2\sqrt{n}}\]

Step 3: Determine convergence of comparison series

The series can be rewritten as: \[\frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}\] This is a constant multiple of a p-series with p = 1/2, which is less than 1. P-series with p ≤ 1 are known to diverge. Therefore, the comparison series diverges.

Step 4: Conclude convergence of the original series

Since we found that a simpler comparison series with terms greater than or equal to the terms of the given series diverges, we can conclude by the Comparison Test that the original series also diverges. So, the given series \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n+1}+\sqrt{n}}\) diverges.