# Thread: Find the convergence of the series

1. ## Find the convergence of the series

Question: Find the convergence of the series

$\displaystyle \frac{1}{1.2.3} + \frac{3}{2.3.4} + \frac{5}{3.4.5} +....$

2. Originally Posted by zorro
Question: Find the convergence of the series

$\displaystyle \frac{1}{1.2.3} + \frac{3}{2.3.4} + \frac{5}{3.4.5} +....$

The general term is $\displaystyle \frac{2n-1}{n(n+1)(n+2)}\le \frac{2n}{n^3}=\frac{2}{n^2}$ , so by comparison...

Tonio

3. ## I didnt understand this

Originally Posted by tonio
The general term is $\displaystyle \frac{2n-1}{n(n+1)(n+2)}\le \frac{2n}{n^3}=\frac{2}{n^2}$ , so by comparison...

Tonio

Could u please tell me how u got this

$\displaystyle \frac{2n-1}{n(n+1)(n+2)}\le \frac{2n}{n^3}$ .........I didnt understand this

4. Man. I gotta know what classes you are taking - you are all over the place!

In any case, when we use the comparison test, we are only interested in the "dominant" terms in the numerator and denominator of our series. For the series you posted (which Tonio translated) the dominant term in the numerator is $\displaystyle 2n$; in the denominator it is $\displaystyle n^3$ (if you multiply out the three terms in the denominator you will get an $\displaystyle n^3$ and some other junk, but it is not needed to solve this problem - all you need to know is the dominant term. Once you determine what your dominant terms are, you can see that that the series you need to compare it to is $\displaystyle \frac{2n}{n^3}$ which simplifies to $\displaystyle \frac{2}{n^2}$ which converges, therefore the series that is LESS than $\displaystyle \frac{2}{n^2}$ must also converge.

Are you in a formal class setting Zorro? The only difficult part of this problem was figuring out the general form of the nth term of the series.

Thank u