Question

How can you show that sum{2^n / (n+1)^n} from n=1->inf is a convergent series using the comparison test

Answer 1

So if we want to show that a series is convergent using the comparison test, then we need another series, which is known to convergent, such that this known convergent series provides an upper bound to our new series.
This is not always so obvious

We need to examine the structure of our series in question in order to postulate a form that would work. If we rewrite the term 2^n / (n+1)^n as (2/(n+1))^n. We can notice that for all n the term is less than or equal to one
This may remind us of geometric series, which we known are convergent to r <1
so let's take out the n=1 term