Test the series for convergence or divergence. \sum_{n=0}^\infty\frac{(-1

Lipossig 2021-10-24 Answered
Test the series for convergence or divergence.
\(\displaystyle{\sum_{{{n}={0}}}^{\infty}}{\frac{{{\left(-{1}\right)}^{{{n}+{1}}}}}{{\sqrt{{{n}+{4}}}}}}\)

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Expert Answer

Raheem Donnelly
Answered 2021-10-25 Author has 20704 answers
Given:
\(\displaystyle{\sum_{{{n}={0}}}^{\infty}}{\frac{{{\left(-{1}\right)}^{{{n}+{1}}}}}{{\sqrt{{{n}+{4}}}}}}\)
On applying radical rule
\(\displaystyle{\sum_{{{n}={0}}}^{\infty}}{\frac{{{\left(-{1}\right)}^{{{n}+{1}}}}}{{\sqrt{{{n}+{4}}}}}}={\sum_{{{n}={0}}}^{\infty}}{\frac{{{\left(-{1}\right)}^{{{n}+{1}}}}}{{{\left({n}+{4}\right)}^{{\frac{{1}}{{2}}}}}}}\)
On applying rule exponent rule
\(\displaystyle{\sum_{{{n}={0}}}^{\infty}}{\frac{{{\left(-{1}\right)}^{{{n}+{1}}}}}{{\sqrt{{{n}+{4}}}}}}={\sum_{{{n}={0}}}^{\infty}}{\left(-{1}\right)}^{{{n}+{1}}}{\left({n}+{4}\right)}^{{{\frac{{-{1}}}{{{2}}}}}}\)
On simplifying,
\(\displaystyle{\left({n}+{4}\right)}^{{\frac{{-{1}}}{{{2}}}}}={\frac{{{1}}}{{\sqrt{{{n}+{4}}}}}}\)
\(\displaystyle{\sum_{{{n}={0}}}^{\infty}}{\frac{{{\left(-{1}\right)}^{{{n}+{1}}}}}{{\sqrt{{{n}+{4}}}}}}={\sum_{{{n}={0}}}^{\infty}}{\left(-{1}\right)}^{{{n}+{1}}}{\frac{{{1}}}{{\sqrt{{{n}+{4}}}}}}\)
On applying the alternating series test the function converges
therefore, \(\displaystyle{\sum_{{{n}={0}}}^{\infty}}{\frac{{{\left(-{1}\right)}^{{{n}+{1}}}}}{{\sqrt{{{n}+{4}}}}}}\) converges
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