Testing convergence of a series using comparison test: $\sum_{k=0}^{\infty} \frac{\sqrt{k+1}}{2^k}$? Can someone please explain to me why this series converged? In my textbook they compare it with geometric series that I don't understand. How am I supposed to come up with this? The series is:
$$\sum_{k=0}^{\infty} \frac{\sqrt{k+1}}{2^k}\tag1$$
They compare it to:
$$\frac{\sqrt{k+1}}{2^k}\leq \left( \frac{2}{3}\right)^k. $$
I understand that this geometric series converges and because of that (1) converges, too.
I just
wonder how should I come up with $\left( \frac{2}{3}\right)^k$?
 A: Basically $k^\alpha\ll c^k$ for $\alpha\ge 0,\ c>1$ provided $k$ is large enough (we are not interested in the first terms anyway for the series convergence). Power functions are dominated by exponentials.
So we get $\dfrac{\sqrt{k+1}}{2^k}\le \dfrac{c^k}{2^k}=\left(\dfrac c2\right)^k$.
For the RHS to converge we need $\dfrac c2<1$, and still $c>1$ thus $1<c<2$.
A simple choice would be $c=\dfrac 32$ but in the present case your textbook did choose $c=\dfrac 43$ which works as well.
A: Since $2^k$ grows horribly faster than $\sqrt{k+1}$, any crude bound does the job pretty nicely.
You may also just use Cauchy-Schwarz:
$$ S=\sum_{k\geq 0}\frac{\sqrt{k+1}}{2^k}\leq \sqrt{\sum_{k\geq 0}\frac{1}{2^k}\sum_{k\geq 0}\frac{k+1}{2^k}} = \sqrt{2\cdot 4}=2\sqrt{2}$$
or something like
$$ S = 1+\sum_{k\geq 1}\frac{\sqrt{k+1}}{2^k} = 1+\sum_{k\geq 0}\frac{\sqrt{k+2}}{2^{k+1}}=1+\frac{S}{2}+\sum_{k\geq 0}\frac{1}{2^k(\sqrt{k+1}+\sqrt{k+2})} $$
and Cauchy-Schwarz to get better bounds. Numerically $S\approx 2.6945075$.
A: Applying the ratio test, we get
$$\frac12\sqrt{\frac{k+2}{k+1}}\le0.61237\cdots\approx\frac23$$ for all $k>0$.
