I am trying to determine if the sum $\sum_{n=1}^\infty \frac{n+4^n}{n+6^n}$ is convergent or divergent.

In attempting to solve this problem I began by seeing if I could do the divergence test - in this case it is not helpful. The terms $4^n,6^n$ look like geometric series type terms but the $n$ in the numerator and denominator are troublesome. I first tried factoring $n$ out of the expression:

$$\sum_{n=1}^\infty \frac{n+4^n}{n+6^n} = \sum_{n=1}^\infty \frac{1+\frac{4^n}{n}}{1+\frac{6^n}{n}}$$

Next I tried comparing to the series $$\sum_{n=1}^\infty \frac{\frac{4^n}{n}}{\frac{6^n}{n}} = \sum_{n=1}^\infty \frac{{4^n}}{{6^n}} = \sum_{n=1}^\infty{\left(\frac{2}{3}\right)}^n$$

This is a convergent geometric series as $r = \frac{2}{3}\lt1$

now I reason that the series $\sum_{n=1}^\infty \frac{1+\frac{4^n}{n}}{1+\frac{6^n}{n}}$ is neither smaller not greater than the series $\sum_{n=1}^\infty{(\frac{2}{3})}^n$ Am I able to conclude anything in this case by a direct comparison test or can I only conclude that the first series is convergent if the second is greater than the first?



Even though you can't do any useful direct comparison with $\left(\frac{2}{3}\right)^n$ because it gives lower bound which converges, you can find upper bound that converges.

In general we would like either lower bound that diverge, or an upper bound that converges. And how to find upper bound for a fraction? Usually by finding upper bound for nominator and lower bound for denominator.

For example $n+4^n < 4^n+ 4^n = 2\cdot 4^n$ and $n+6^n>6^n$, so

$$ \frac{4^n+n}{6^n+n} < \frac{ 2\cdot 4^n}{6^n} = 2 \left(\frac{2}{3}\right)^n. $$

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    $\begingroup$ Thanks @Sil, i like this answer. That's a really neat trick i haven't seen before! $\endgroup$ – Blargian Nov 25 '17 at 7:17

By the direct comparison test, you cannot conclude anything, as

$$\frac{4^n+n}{6^n+n} > \frac{4^n}{6^n}.$$

However, you can use the limit comparison test, which will allow you to deduce the result.

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    $\begingroup$ You could also use direct comparison by noting that $$\frac{n+4^n}{n+6^n} \le \frac{4^n + 4^n}{6^n} = 2 \left( \frac 23 \right)^n$$ $\endgroup$ – User8128 Nov 25 '17 at 7:12
  • $\begingroup$ I thought of that however i wasn't sure what i should take as my $b_n$ term? If i factor out the $n$ on the top and the bottom would i take $b_n$ as $\frac{1}{1+\frac{6^n}{n}}$? $\endgroup$ – Blargian Nov 25 '17 at 7:12
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    $\begingroup$ @Blargian Since the things you're summing are $\frac{4^n+n}{6^n+n}$ and $\frac{4^n}{6^n}$, simply pick one of them to be $a_n$ and the other to be $b_n$. $\endgroup$ – Carl Schildkraut Nov 25 '17 at 7:18
  • $\begingroup$ @Carl Schildkraut I take $b_n = \frac{4^n}{6^n}$ then $\lim_{n \to \infty}\frac{a_n}{b_n} = \lim_{n \to \infty}\frac{4^n+n}{6^n+n} \frac{6^n}{4^n}$ and i simplify this to $\frac{1+\frac{n}{4^n}}{1+\frac{n}{6^n}}$ I am unsure of how to proceed with finding this limit. I know that if the result is $\gt 1$ then the series will converge. $\endgroup$ – Blargian Nov 25 '17 at 7:32
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    $\begingroup$ With the limit comparison test, if the limit is any positive real number, then the series either both converge or both diverge. In this case, you’ll get a limit of $1$, which means the test series converges, just like the known series. You get $1$ by noting that both $\frac{n}{4^n}$ and $\frac{n}{6^n}$ go to $0$ as $n\to\infty$. $\endgroup$ – G Tony Jacobs Nov 25 '17 at 7:59

One way to use direct comparison is this:

$$\frac{4^n+n}{6^n+n}\le \frac{5^n}{6^n}=\left(\frac56\right)^n.$$


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