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

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

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Raheem Donnelly
Given:
$$\displaystyle{\sum_{{{n}={0}}}^{\infty}}{\frac{{{\left(-{1}\right)}^{{{n}+{1}}}}}{{\sqrt{{{n}+{4}}}}}}$$
$$\displaystyle{\sum_{{{n}={0}}}^{\infty}}{\frac{{{\left(-{1}\right)}^{{{n}+{1}}}}}{{\sqrt{{{n}+{4}}}}}}={\sum_{{{n}={0}}}^{\infty}}{\frac{{{\left(-{1}\right)}^{{{n}+{1}}}}}{{{\left({n}+{4}\right)}^{{\frac{{1}}{{2}}}}}}}$$
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