Explained: "Test the series for convergence or divergence. ∑n=0∞(−1)n+1n+4"

Lipossig

Answered question

2021-10-24

Test the series for convergence or divergence.

\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n + 1}}{\sqrt{n + 4}}

Raheem Donnelly

Skilled2021-10-25Added 75 answers

Given:

\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n + 1}}{\sqrt{n + 4}}

On applying radical rule

\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n + 1}}{\sqrt{n + 4}} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n + 1}}{{(n + 4)}^{\frac{1}{2}}}

On applying rule exponent rule

\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n + 1}}{\sqrt{n + 4}} = \sum_{n = 0}^{\infty} {(- 1)}^{n + 1} {(n + 4)}^{\frac{- 1}{2}}

On simplifying,

{(n + 4)}^{\frac{- 1}{2}} = \frac{1}{\sqrt{n + 4}}

\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n + 1}}{\sqrt{n + 4}} = \sum_{n = 0}^{\infty} {(- 1)}^{n + 1} \frac{1}{\sqrt{n + 4}}

On applying the alternating series test the function converges
therefore,

\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n + 1}}{\sqrt{n + 4}}

converges

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