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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

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    I’m pretty sure that @DonAntonio’s edit is correct, but you didn’t use enough parentheses to make your expressions unambiguous, and you had a couple of equals signs that made no sense, so do please check.2012-09-28

2 Answers 2

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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} $

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    Of course, that'll make it.2012-09-28
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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.

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    Nice addition! $\quad +1\quad \ddot\smile \quad \checkmark!\;$2013-03-23