Question

Check the convergence$\sum _{k=1}^{\infty }\frac{1}{k+\sqrt{k}}$

Short answer

Diverges

Step-by-step solution

Step 1. Given 

The given series is$\sum _{k=1}^{\infty }\frac{1}{k+\sqrt{k}}$

Step 2. Limit comparison test

$$\begin{gathered} \mathrm{According}\mathrm{to}\mathrm{limit}\mathrm{comparison}\mathrm{test},\mathrm{let} \\ \sum _{k=1}^{\infty }{a}_{k}and\sum _{k=1}^{\infty }{b}_{\mathrm{k}}\mathrm{be}\mathrm{two}\mathrm{series}\mathrm{with}\mathrm{positive}\mathrm{terms} \\ 1.\mathrm{If}\underset{\mathrm{k}\to \infty }{\lim }\frac{{\mathrm{a}}_{\mathrm{k}}}{{\mathrm{b}}_{\mathrm{k}}}=\mathrm{L},\mathrm{where}\mathrm{L}\mathrm{is}\mathrm{any}\mathrm{positive}\mathrm{real}\mathrm{number}, \\ \mathrm{then}\mathrm{either}\mathrm{both}\mathrm{series}\mathrm{converges}\mathrm{or}\mathrm{both}\mathrm{series}\mathrm{diverges}. \\ 2.\underset{\mathrm{k}\to \infty }{\lim }\frac{{\mathrm{a}}_{\mathrm{k}}}{{\mathrm{b}}_{\mathrm{k}}}=0,\sum _{\mathrm{k}=1}^{\infty }{\mathrm{b}}_{\mathrm{k}}\mathrm{converges}\mathrm{then}\sum _{\mathrm{k}=1}^{\infty }{\mathrm{a}}_{\mathrm{k}}\mathrm{diverges} \\ 3.\underset{\mathrm{k}\to \infty }{\lim }\frac{{\mathrm{a}}_{\mathrm{k}}}{{\mathrm{b}}_{\mathrm{k}}}=\infty ,\sum _{\mathrm{k}=1}^{\infty }{\mathrm{b}}_{\mathrm{k}}\mathrm{diverges}\mathrm{then}\sum _{\mathrm{k}=1}^{\infty }{\mathrm{a}}_{\mathrm{k}}\mathrm{Converges} \end{gathered}$$

Step 3. Checking the convergence 

$$\begin{gathered} a_k=\frac{1}{k+\sqrt{k}}and{b}_k=\frac{1}{k} \\ \underset{k\to \infty }{\lim }\frac{a_k}{b_k}=\underset{k\to \infty }{\lim }\frac{\frac{1}{k+\sqrt{k}}}{\frac{1}{k}} \\ s\mathrm{olving}\mathrm{this}\mathrm{limit}=\frac{1}{1+0} \\ =1 \\ \sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{2^k} \\ \mathrm{The}\mathrm{given}\mathrm{serie}s\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k}\mathrm{is}a\mathrm{harmonic}\mathrm{series},\mathrm{hence}\mathrm{diverges} \\ \mathrm{From}\mathrm{limit}\mathrm{comparison}\mathrm{test}\mathrm{the}\mathrm{series}\mathrm{Diverges} \end{gathered}$$