The given series is alternating, and $\underset{n\to \infty }{lim}1\/\sqrt{n\+1}$ = $0$, and the sequence $\left\{1\/\sqrt{n\+1}\right\}$  is monotone decreasing, with limit zero.

Hence, by the Leibniz test, the series converges conditionally.

However, it does not converge absolutely, as can be seen via the Limit-Comparison test, using the divergent p-series $\mathrm{\&Sigma;} 1\&sol;\sqrt{n}$ ($p\&equals;1\&sol;2<1$):

$\underset{n\to \infty }{lim}\frac{1\&sol;\sqrt{n\&plus;1}}{1\&sol;\sqrt{n}}$ = $\underset{n\to \infty }{lim}\sqrt{\frac{n}{n\&plus;1}}\&equals;1$

By part (1) of the Limit-Comparison test, both series will converge or diverge, and since the $p$-series diverges, so also does the given series if all its terms are positive.$$