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I have the following problem

$A=\sum_{n=1}^{\infty}\frac{2n+7^n}{2n+6^n}$

... and I'm trying to figure out if it is convergent or divergent using the Comparison Test.


A similar problem:

$D = \sum_{n=1}^{\infty}\frac{4n+5^n}{4n+8^n}$

... is solved by using the Comparison Test like so:

$\frac{4n+5^n}{4n+8^n}<\frac{5^n+5^n}{4n+8^n}<\frac{2\cdot 5^n}{8^n}$

and since:

$2\sum _{n=1}^{\infty }\frac{5^n}{8^n}$

... is a convergent Geometric Series, $D$ converges too.


If I try to do something similar to the original problem, $A$, I get the following:

$B=\sum _{n=1}^{\infty }\frac{7^n}{6^n}$

... is a divergent geometric series.

Therefore, I believe I need to use the comparison test to show that:

$A>B$

... this time, and I will be able to conclude that $A$ is divergent as well. If I try doing that, though, I run into a problem:

$\frac{2n+7^n}{2n+6^n}>\frac{7^n}{2n+6^n}<\frac{7^n}{6^n}$

Am I comparing it to the wrong function? Can I not use the Comparison Test on this problem and instead need to use the Integral Comparison Test?

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    In other words, you can do comparison \frac{2n+7^n}{2n+6^n}>1 and get divergence.2012-03-22

3 Answers 3

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As Andre Nicolas said your terms do not approach zero, and hence your series will diverge.

If however you want to use the comparison test you are pretty close $\frac{2n+7^n}{2n+6^n}>\frac{7^n}{2n+6^n}>\frac{7^n}{6^n+6^n}=\frac{1}{2}\cdot\frac{7^n}{6^n}$ And so your since $\frac{7}{6}>1$, you can say the series of the right diverge, and hence your series is larger than a divergent, and hence divergent.

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If you really want to do a comparison of the type you described, note that $2n+6^n<6^n+6^n$. So the terms of your series are greater than $\frac{7^n}{2\cdot 6^n}.$

Remark: But this is too much work. Look at the original series. The top is always bigger than the bottom.

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    Once again I apologise, i$f$ it came across to you wrongly. I understand your perspective too. : )2012-03-22
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One relation… $\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$