Using the Limit Comparison Test

Determine whether the series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{2k^{3/2}+5}\) converges or diverges.

Solution We choose an appropriate \(p\)-series to use for comparison by examining the behavior of the series for large values of \(n\): \[ \begin{eqnarray*} \dfrac{1}{2n^{3/2}+5}=\dfrac{1}{n^{3/2}\left( 2+\dfrac{5}{n^{3/2}}\right) }= \dfrac{1}{n^{3/2}}\left( \dfrac{1}{2+\dfrac{5}{n^{3/2}}}\right) & \underset{\underset{\color{#0066A7}{{\hbox{for large }\hbox{\(n\)}}}}{{\color{#0066A7}{{\uparrow }}}} }{\approx} & \dfrac{1}{n^{3/2}}\left(\dfrac{1}{2}\right)\\ \end{eqnarray*} \]

This leads us to choose the \(p\)-series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k^{3/2}},\) which converges, and use the Limit Comparison Test with \[ a_{n}=\dfrac{1}{2n^{3/2}+5} \qquad \hbox{and}\qquad b_{n}=\dfrac{1}{n^{3/2}} \] \[ \lim\limits_{n\rightarrow \infty }\dfrac{a_{n}}{b_{n}}=\lim_{n\,\rightarrow \,\infty }\frac{\dfrac{1}{2n^{3/2}+5}}{\dfrac{1}{n^{3/2}}} =\lim\limits_{n\rightarrow \infty }\frac{n^{3/2}}{2n^{3/2}+5} =\lim\limits_{n\rightarrow \infty }\dfrac{1}{2+\dfrac{5}{n^{3/2}}}=\frac{1}{2} \]

Since the limit is a positive number and the \(p\)-series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k^{3/2}}\) converges, then by the Limit Comparison Test, \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{2k^{3/2}+5}\) also converges.