# Is my interpretation of direct comparison test correct?

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 ?

• $\frac {1}{2^n + n} < \frac {1}{2^n}.$ To use the comparison test we must show that partial sums of a series are always less than the partial sums of a series that we know to be convergent. If partial sums are greater, we can use that to prove that a series diverges. Commented May 9, 2019 at 5:47

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.
• Does that mean we should find a series that converges ($b_n$) with $a_n \le b_n$ holding true and then that guarantees convergence of $a_n$? Commented May 9, 2019 at 5:31
• @DeepamSarmah If $\sum a_n$ converges, then finding $b_n \ge a_n$ such that $\sum b_n$ converges will show convergence of $\sum a_n$. If $\sum a_n$ does not converge, then obviously no such choice of $b_n$ will exist. Conversely, if $\sum a_n$ absolutely diverges, then finding $b_n \le a_n$ such that $\sum b_n$ diverges will show $\sum a_n$ diverges. Commented May 9, 2019 at 5:34
• so it means that if $\Sigma a_n$ converges then there will exist $\Sigma b_n$ that will also converge with $a_n \le b_n$? Commented May 9, 2019 at 5:43
• @DeepamSarmah I am not aware of a theorem that ensures the existence of such a choice for any $a_n$, so I cannot tell you if this is true or false. I can say, however, that in practice, it is not always practical to find a $b_n$ for which $b_n \ge a_n$ for all $n$ and it is "easier" to show $\sum b_n$ converges than to show $\sum a_n$ converges directly. Commented May 9, 2019 at 5:55