I had been studying convergence and divergence of series in Calculus. This is the problem that I've to prove using direct comparison test.
Prove that the series $\sum_{n=1}^{\infty} \frac{1}{2^n + n}$ converges.
My solution goes like this, we know that $\frac{1}{2^n} > \frac{1}{2^n + n} \; \forall \; n \in \mathbb{N}$ and so by proving convergence of $\sum_{n=1}^{\infty} \frac{1}{2^n}$ we can prove that $\sum_{n=1}^{\infty} \frac{1}{2^n + n}$ converges.
But another way to interpret is to consider that $\frac{1}{n} >\frac{1}{2^n + n} $. But now $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges and so we cannot apply direct comparison test here.
My question is we can have many $b_n$ such that $a_n \le b_n$ ($a_n$ is the series which we want to test for convergence) so for applying direct comparison test should we have at least $1$ $b_n$ (say $c_n$) such that $\sum_{n = 1}^{\infty}c_n$ converges ?
The choice of comparison is important, because I can of course compare $$1 > \frac{1}{2^n + n},$$ or for something even more absurd, $$10^{10^{10}} > \frac{1}{2^n+n}.$$ The point of bounding the summand from above is to show that there exists a choice for which the sum still converges and dominates the original series; you don't--and in fact, cannot--show that any choice still converges.