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

Solution 1:

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

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

Solution 2:

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.

Solution 3:

One way to use direct comparison is this:

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