Don't understand this convergence/divergence problem?

• Nov 18th 2010, 06:53 PM
Reefer
Don't understand this convergence/divergence problem?
It's the sum from n=1 to infinity of 4/(n^2+2n).

Shouldn't this be converging to 0? I don't understand how to work this out and I have my midterm tomorrow. It says it converges to 3 but I don't understand how it gets there. What convergence test do I use?
• Nov 18th 2010, 07:29 PM
Prove It
Try a partial fraction decomposition.

$\displaystyle \frac{A}{n} + \frac{B}{n + 2} = \frac{4}{n(n + 2)}$

$\displaystyle \frac{A(n + 2) + Bn}{n(n + 2)} = \frac{4}{n(n + 2)}$

$\displaystyle A(n + 2) + Bn = 4$

$\displaystyle An + 2A + Bn = 4$

$\displaystyle (A+B)n + 2A = 0n + 4$

$\displaystyle A+B = 0$ and $\displaystyle 2A = 4$

$\displaystyle A = 2$ and $\displaystyle B = -2$.

Therefore $\displaystyle \frac{2}{n} - \frac{2}{n + 2} = \frac{4}{n(n + 2)}$.

So $\displaystyle \sum_{n = 1}^{\infty}\frac{4}{n^2 + 2n} = \lim_{m \to \infty}\sum_{n = 1}^m\left(\frac{2}{n} - \frac{2}{n + 2}\right)$

$\displaystyle = \lim_{m \to \infty}\left[\left(\frac{2}{1} - \frac{2}{3}\right) + \left(\frac{2}{2} - \frac{2}{4}\right) + \left(\frac{2}{3} - \frac{2}{5}\right) + \left(\frac{2}{4} - \frac{2}{6}\right) + \dots\right$
$\displaystyle\left + \left(\frac{2}{m-3} - \frac{2}{m-1}\right) + \left(\frac{2}{m-2} - \frac{2}{m}\right) + \left(\frac{2}{m-1} - \frac{2}{m+1}\right) + \left(\frac{2}{m} - \frac{2}{m + 2}\right)\right]$

$\displaystyle =\lim_{m \to \infty}\left(\frac{2}{1} + \frac{2}{2} - \frac{2}{m + 1} - \frac{2}{m + 2}\right)$

$\displaystyle = 3$.

Clearly this sum is convergent and converges to $\displaystyle 3$.
• Nov 18th 2010, 08:42 PM
Soroban
Hello, Reefer!

Quote:

$\displaystyle S \;=\;\sum ^{\infty}_{n = 1} \frac{4}{n^2+2n}$

Shouldn't this be converging to 0?
No, the terms are getting smaller and converge to 0,

It says it converges to 3, but I don't understand how it gets there.

We have: . $\displaystyle S \;=\; 4\sum^{\infty}_{n=1}\frac{1}{n(n+2)}$

Partial Fraction Decomposition

We have: . $\dfrac{1}{n(n+2)} \:=\:\dfrac{A}{n} + \dfrac{B}{n+2}$

Multiply through by $n(n+2)\!:\;\;1 \;=\;A(n+2) + B(n)$

. . Let $n = 0\!:\;\;1 \:=\:A(0+2) + B(0) \quad\Rightarrow\quad A \:=\:\frac{1}{2}$

. . Let $n = -2:\;\;1 \:=\:A(0) + B(-2) \quad\Rightarrow\quad B \:=\:-\frac{1}{2}$

Hence: . $\displaystyle \frac{1}{n(n+2)} \;=\;\frac{\frac{1}{2}}{n} + \frac{-\frac{1}{2}}{n+2} \;=\;\frac{1}{2}\left(\frac{1}{n} - \frac{1}{n+2}\right)$

Then: . $\displaystyle S \;=\;4\sum^{\infty}_{n=1}\frac{1}{2}\left(\frac{1} {n} - \frac{1}{n+2}\right) \;=\;2\sum^{\infty}_{n=1}\left(\frac{1}{n} - \frac{1}{n+2}\right)$

Write out the infinite sum:

$\displaystyle S \;=\;2\bigg[\left(\frac{1}{1} - \frac{1}{3}\right) + \left(\frac{1}{2}-\frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \left(\frac{1}{4} - \frac{1}{6}\right) + \hdots \bigg]$

We see that all the fractions cancel out except $\frac{1}{1}$ and $\frac{1}{2}$

So we have: . $S \;=\;2\left(\frac{1}{1} + \frac{1}{2}\right) \;=\;2\left(\frac{3}{2}\right)$

. .Therefore: . $S \:=\:3$

• Nov 19th 2010, 03:08 AM
HallsofIvy
Quote:

Originally Posted by Reefer
It's the sum from n=1 to infinity of 4/(n^2+2n).

Shouldn't this be converging to 0?

The sum obviously doesn't converge to 0 since it is a sum of positive numbers!
$\frac{4}{1+ 2}+ \frac{4}{4+ 4}+ \frac{4}{9+ 5}+ \cdot\cdot\cdot= \frac{4}{3}+ \frac{1}{2}+ \frac{2}{7}+ \cdot\cdot\cdot$
couldn't possibly be 0!

Are you confusing the series (sum) with the sequence? In order that a series converge, the sequence of values must converge to 0.

Quote:

I don't understand how to work this out and I have my midterm tomorrow. It says it converges to 3 but I don't understand how it gets there. What convergence test do I use?