Use the Comparsion Test or Limit Comparsion Test to determine whether the series converges or not.
$\sum^\infty_{n=1} \frac{n+\sqrt n}{n+n^2}$
$\sum^\infty_{n=1}\frac{1+3^n}{1+2^n}$
thanks a lot
Use the Comparsion Test or Limit Comparsion Test to determine whether the series converges or not.
$\sum^\infty_{n=1} \frac{n+\sqrt n}{n+n^2}$
$\sum^\infty_{n=1}\frac{1+3^n}{1+2^n}$
thanks a lot
Hints (pretty big, though, and assuming my editing the question is correct):
$(1)\;\;\;\;\;\;\;\;\;\frac{n+\sqrt n}{n+n^2}\geq \frac{n}{n^2+n^2}$
$(2)\;\;\;\;\;\;\;\;\;\frac{1+3^n}{1+2^n}\geq \frac{ 3^n}{2\cdot 2^n} $
The way @DonAntonio did above completes the proof. I just add another small approach to the second series. I use the ratio test for it so I am looking for the following limit when $n\to\infty$: $\lim\bigg|\frac{a_{n+1}}{a_n}\bigg|$ where in $a_n=\frac{1+3^n}{1+2^n}$. So I have: $\lim\bigg|\frac{\frac{1+3^{n+1}}{1+2^{n+1}}}{\frac{1+3^n}{1+2^n}}\bigg|=\lim\bigg|0.5\bigg(\frac{1}{1+2^{n+1}}+1\bigg)\bigg(\frac{3}{\frac{2}{1+3^{n+1}}+1}\bigg)\bigg|=\frac{3}{2}>1$ when $n$ tends to infinity. Therefore the second one is divergent.