Using the Limit Comparison Test

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

Solution We choose a \(p\)-series for comparison by examining how the terms of the series behave for large values of \(n\): \[ \begin{eqnarray*} \frac{3\sqrt{n}+2}{\sqrt{n^{3}+3n^{2}+1}}=\frac{\sqrt{n}\,\left( 3+\dfrac{2}{ \sqrt{n}}\right) }{\sqrt{n^{3}}\left( \sqrt{1+\dfrac{3}{n}+\dfrac{1}{n^{3}}} \right) }=\dfrac{n^{1/2}}{n^{3/2}}\frac{3+\dfrac{2}{\sqrt{n}}}{\sqrt{1+ \dfrac{3}{n}+\dfrac{1}{n^{3}}}} & \underset{\underset{\color{#0066A7}{{\hbox{for large }\hbox{\(n\)}}}}{{\color{#0066A7}{{\uparrow}}}}}{\approx} & \dfrac{1}{n}(3) \\ \end{eqnarray*} \]

So, we compare the series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{3\sqrt{k~}+2}{\sqrt{k^{3}+3k^{2}+1}}\) to the harmonic series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k}\), which diverges, and use the Limit Comparison Test with \[ a_{n}=\frac{3\sqrt{n}+2}{\sqrt{n^{3}+3n^{2}+1}}\qquad \hbox{ and }\qquad b_{n}=\dfrac{1}{n} \] \[ \begin{eqnarray*} \lim\limits_{n\rightarrow \infty }\dfrac{a_{n}}{b_{n}} &=& \lim_{n\,\rightarrow \,\infty }\frac{\dfrac{3\sqrt{n}+2}{\sqrt{n^{3}+3n^{2}+1}}}{\dfrac{1}{n}} =\lim\limits_{n\rightarrow \infty }\frac{n\left( 3\sqrt{n}+2\right) }{\sqrt{ n^{3}+3n^{2}+1}}=\lim\limits_{n\rightarrow \infty }\frac{3n^{3/2}+2n}{\sqrt{n^{3}+3n^{2}+1}}\\[7pt] &=&\lim\limits_{n\rightarrow \infty }\dfrac{3n^{3/2}}{n^{3/2}}=3 \end{eqnarray*} \]

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