Hello, guys!
I have two series: $\displaystyle \sum_{n=1}^{\infty}(-1)^{n}\dfrac{3n-1}{n^2 + n}$
My question is how to figure it out their convergence / divergence.
Hello, guys!
I have two series: $\displaystyle \sum_{n=1}^{\infty}(-1)^{n}\dfrac{3n-1}{n^2 + n}$
My question is how to figure it out their convergence / divergence.
The first is an alternating series with terms that decrease to 0. Therefore it is convergent.
For the second series, I see that "$\displaystyle \left(\frac{1}{2}\right)^n$ and immediately think "compare it to a geometric sequence". First, write it as $\displaystyle \frac{ln\left(\frac{n+ 3}{n}\right)}{3}\left(\frac{1}{2}\right)^n$
$\displaystyle \frac{ln\left(\frac{n+ 3}{n}\right)}{3}< 1$ if and only if $\displaystyle ln\left(\frac{n+ 3}{n}\right)< 3$ which is true if and only if $\displaystyle \frac{n+ 3}{n}< e^3$. $\displaystyle n+ 3< e^3n$ if and only if $\displaystyle n(1- e^3)< 3$ or $\displaystyle (e^3- 1)n> 3$. That is, each term is less than $\displaystyle \left(\frac{1}{2}\right)^n$. The series has the convergent geometric series as upper bound so converges.