# Thread: How to solve two series to figure it out the convergence / divergence

1. ## How to solve two series to figure it out the 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.

2. ## Re: How to solve two series to figure it out the convergence / divergence

The second one is $$\sum_{n=1}^{\infty}\dfrac{\ln\left(\dfrac{n+3}{n }\right)}{3 \cdot 2^{n}}$$.

3. ## Re: How to solve two series to figure it out the 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.

4. ## Re: How to solve two series to figure it out the convergence / divergence

@HallsofIvy, can you show that first one is decrease to $0$.

5. ## Re: How to solve two series to figure it out the convergence / divergence

Originally Posted by lebdim
@HallsofIvy, can you show that first one is decrease to $0$.
Surely you see that $\displaystyle \mathop {\lim }\limits_{n \to \infty } \dfrac{{3n - 1}}{{{n^2} + n}} = 0$ is true?

Can you use a derivative to show that $\dfrac{3x-1}{x^2+x}$ is a decreasing function?