Problem 26 - Calculus: One and Several Variables

Question

Determine whether the series converges or diverse. $$\sum \frac{2 k+1}{\sqrt{k^{3}+1}}$$

Step-by-Step Solution

Answer

The series diverges.

Step 1: Identify Test

Because the series possesses both positive terms and an apparent end behavior, the Comparison Test seems applicable.

Step 2: Identify Comparable series

Next is to find a series to compare. In this case, if you simplify the terms of the series by removing lower order terms, you get \( a_n = \frac{2n}{n^{3/2}} = \frac{2}{n^{1/2}} \). So it is cleared that the series \( \sum \frac{2}{n^{1/2}} \) can be a good series to compare the given series to.

Step 3: Apply Comparison Test Criteria

For all \( k \geq 1 \) in the function, \( \frac{2k+1}{\sqrt{k^{3}+1}} \leq \frac{2k}{k^{3/2}} = \frac{2}{k^{1/2}} \) which means that for all \( k \geq 1 \), the terms of the given series are less than or equal to the terms in the series \( \sum \frac{2}{k^{1/2}} \) which is a p-series with \( p=1/2 \). Considering that p-series will only converge if \( p > 1 \) and here \( p=1/2<1 \). So, the comparison series diverges.

Step 4: Conclude if Series Converges or Diverges

Since the comparison series diverges according to the Comparison Test, the original series \( \sum \frac{2k+1}{\sqrt{k^{3}+1}} \) also diverges.