We have an answer to "Test the convergence of the following series: n+1−1(n+2)3−1+…+∞"

jubateee

Answered question

2022-01-07

Test the convergence of the following series:

\frac{\sqrt{n + 1} - 1}{{(n + 2)}^{3} - 1} + \dots + \infty

Answer & Explanation

limacarp4

Beginner2022-01-08Added 39 answers

Since

\frac{\sqrt{n + 1} - 1}{{(n + 2)}^{3} - 1} = \frac{1}{n^{\frac{5}{2}}} \frac{\sqrt{1 + \frac{1}{n}} - \sqrt{\frac{1}{n}}}{{(1 + \frac{2}{n})}^{3} - \frac{1}{n^{3}}}

\sim \frac{1}{n^{\frac{5}{2}}}

your method looks fine.

Ethan Sanders

Beginner2022-01-09Added 35 answers

\frac{\sqrt{n + 1} - 1}{{(n + 2)}^{3} - 1} = \frac{(\sqrt{n + 1} - 1) (\sqrt{n + 1}} + 1}{({(n + 2)}^{3} - 1) (\sqrt{n + 1} + 1)}

\frac{n}{({(n + 2)}^{3} - 1) (\sqrt{n + 1} + 1)} < \frac{n}{n^{3}}

= \frac{1}{n^{2}}

Using the p-test,

\frac{1}{n^{2}}

converges, and since

0 < \frac{n}{({(n + 2)}^{3} - 1) (\sqrt{n + 1} + 1)} < \frac{1}{n^{2}}

, using the comparison test, the original series must also converge.

karton

Expert2022-01-11Added 613 answers

It is clear that $\frac{\sqrt{n + 1} - 1}{(n + 2)^{3} - 1} \sim \frac{1}{n^{\frac{5}{2}}}$ as $n \to \infty$ . Since the series $\sum_{n = 1}^{\infty} \frac{1}{n^{2}}$ converges so that the series must be converge

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