# Convergent/Divergent Series help!

• Apr 26th 2010, 10:35 AM
amywilliams99
Convergent/Divergent Series help!
I have a couple of Q's I'm stuck on help and help would be MUCH appreciated! I have to use either the limit comparison or the comparison test only -
Sum (n=1 to infinity) n/(2^n)
Sum (n=1 to infinity) 1/(2 + n^0.5)
• Apr 26th 2010, 10:54 AM
Failure
Quote:

Originally Posted by amywilliams99
I have a couple of Q's I'm stuck on help and help would be MUCH appreciated! I have to use either the limit comparison or the comparison test only -
Sum (n=1 to infinity) n/(2^n)
Sum (n=1 to infinity) 1/(2 + n^0.5)

Consider:
$\displaystyle \frac{n}{2^n}=\frac{n}{\sqrt{2}^n}\cdot\frac{1}{\s qrt{2}^n}$
Now I assume that you know that $\displaystyle \frac{n}{\sqrt{2}^n}\rightarrow 0$, for $\displaystyle n\rightarrow \infty$, because $\displaystyle \sqrt{2}>1$.
Thus, there exists an $\displaystyle n_0\in\mathbb{N}$, such that $\displaystyle \frac{n}{\sqrt{2}^n}<1$, for all $\displaystyle n>n_0$, and thus the tail of your first series can be compared with the tail of the geometric serie $\displaystyle \sum_{n=n_0+1}\frac{1}{\sqrt{2}^n}<\infty$.

As to your second series my guess is that for sufficiently large $\displaystyle n$, we will have that $\displaystyle \frac{1}{2+\sqrt{n}}\geq \frac{1}{n}$, thus you can compare it with the divergent harmonic series.
To check whether $\displaystyle \frac{1}{2+\sqrt{n}}\geq \frac{1}{n}$ holds true for sufficiently large $\displaystyle n$, you just note that $\displaystyle n\geq 2\sqrt{n}\geq 2+\sqrt{n}$, if $\displaystyle n\geq 4$.