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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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    $\begingroup$ Show what you did first. $\endgroup$ – Ram Sep 28 '12 at 3:53
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    $\begingroup$ Check my editing your series is correct, and please" next time please do read first in our site's FAQ how to correctly write mathematics here, and check that what you wrote makes sense before you post it. $\endgroup$ – DonAntonio Sep 28 '12 at 3:53
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    $\begingroup$ 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. $\endgroup$ – Brian M. Scott Sep 28 '12 at 3:55
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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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  • $\begingroup$ Of course, that'll make it. $\endgroup$ – DonAntonio Sep 28 '12 at 4:31
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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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  • $\begingroup$ Nice addition! $\quad +1\quad \ddot\smile \quad \checkmark!\;$ $\endgroup$ – Namaste Mar 23 '13 at 0:43

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