Don't understand this convergence/divergence problem?

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November 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?
November 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 $\displaystyle 3$ .
November 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,
but they are asking about the sum of the terms.

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: . $S \:=\:3$
November 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?