Although $\frac{1}{2\&plus;\sqrt{n}}\to 0$ as $n\to \infty$, $\sqrt{n}<n$ so that the denominator is smaller than $n$, the denominator of the harmonic series, and hence, the terms of this series will be greater than the terms of the divergent harmonic series. This observation suggests the Limit-Comparison test, with the harmonic series as the comparison series. The relevant calculation is then

$\underset{n\to \infty }{lim}\frac{\left(\frac{1}{2\&plus;\sqrt{n}}\right)}{1\&sol;n}\&equals;\underset{n\to \infty }{lim}\frac{n}{2\&plus;\sqrt{n}}\&equals;\infty$ 

By part (3) of the Limit-Comparison test , if the comparison series diverges, so also the given series diverges.

Alternatively, part (1) of the Limit-Comparison test with the divergent $p$-series $\mathrm{\&Sigma;} 1\&sol;\sqrt{n}$ ($p\&equals;1\&sol;2<1$) as the comparison series also establishes the divergence of the given series because $\underset{n\to \infty }{lim}\frac{\frac{1}{2\&plus;\sqrt{n}}}{1\&sol;\sqrt{n}}$ = $1$.