Answer
Divergent
Work Step by Step
\[\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{13}+\cdots=\sum_{n=1}^{\infty}\frac{1}{2n+3}\]
Let us assume that, \[\sum_{n=1}^{\infty}\frac{1}{2n+3}=\sum_{n=1}^{\infty}a_n\]
\[\Rightarrow a_n=\frac{1}{2n+3}\]
$a_n$ is positive term and montonically decreasing.
So Integral test is applicable.
Let\[I=\int_{1}^{\infty}\frac{1}{2x+3}dx\]
\[I=\lim_{t\rightarrow\infty}\int_{1}^{t}\frac{1}{2x+3}dx\;\;\;\;\;\;\;\ldots(1)\]
Let \[I_1=\int\frac{1}{2x+3}dx\]
Substitute \[y=2x+3\;\;\Rightarrow\;dy=2dx\;\Rightarrow dx=\frac{1}{2}dy\]
\[\Rightarrow I_1=\frac{1}{2}\int \frac{dy}{y}=\frac{1}{2}\ln |y|\]
\[\Rightarrow I_1=\frac{1}{2}\ln |2x+3|\;\;\;\;\;\ldots (2)\]
Using (2) in (1)
\[I=\frac{1}{2}\lim_{t\rightarrow\infty}[\ln |2x+3|]_{x=1}^{t}\]
\[I=\frac{1}{2}\lim_{t\rightarrow\infty}(\ln |2t+3|-\ln |5|)\]
\[I=\infty\]
$I$ is divergent.
By integral test, $\displaystyle\sum_{n=1}^{\infty}\frac{1}{2n+3}$ is divergent.
Hence, $\displaystyle{\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{13}+\cdots}$ is divergent
