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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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  • $\begingroup$ The terms of the series do not approach $0$, so you have automatic divergence. It is important to take a good informal look at a series first. $\endgroup$ Mar 22 '12 at 19:16
  • $\begingroup$ In other words, you can do comparison $$\frac{2n+7^n}{2n+6^n}>1$$ and get divergence. $\endgroup$
    – GEdgar
    Mar 22 '12 at 19:20
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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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  • $\begingroup$ I am sorry to be nitpicky-but we would be sending wrong signals by using top and bottom. Why don't we stick to numerator and denominator? $\endgroup$
    – user21436
    Mar 22 '12 at 19:41
  • $\begingroup$ @KannappanSampath: The use of "top" and "bottom" is admittedly very casual. The idea is to stress, through the use of casual language, that we need to look first at what's happenin'. Students, even very good ones, all too often attack a problem by looking through the screwdrivers in the toolbox. But I will demote the last sentence of the post to a remark. $\endgroup$ Mar 22 '12 at 19:59
  • $\begingroup$ Once again I apologise, if it came across to you wrongly. I understand your perspective too. : ) $\endgroup$
    – user21436
    Mar 22 '12 at 20:00
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One relation… $$\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$$

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