Printable View
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?
Try a partial fraction decomposition.
$\displaystyle \displaystyle \frac{A}{n} + \frac{B}{n + 2} = \frac{4}{n(n + 2)}$
$\displaystyle \displaystyle \frac{A(n + 2) + Bn}{n(n + 2)} = \frac{4}{n(n + 2)}$
$\displaystyle \displaystyle A(n + 2) + Bn = 4$
$\displaystyle \displaystyle An + 2A + Bn = 4$
$\displaystyle \displaystyle (A+B)n + 2A = 0n + 4$
$\displaystyle \displaystyle A+B = 0$ and $\displaystyle \displaystyle 2A = 4$
$\displaystyle \displaystyle A = 2$ and $\displaystyle \displaystyle B = -2$.
Therefore $\displaystyle \displaystyle \frac{2}{n} - \frac{2}{n + 2} = \frac{4}{n(n + 2)}$.
So $\displaystyle \displaystyle \sum_{n = 1}^{\infty}\frac{4}{n^2 + 2n} = \lim_{m \to \infty}\sum_{n = 1}^m\left(\frac{2}{n} - \frac{2}{n + 2}\right)$
$\displaystyle \displaystyle = \lim_{m \to \infty}\left[\left(\frac{2}{1} - \frac{2}{3}\right) + \left(\frac{2}{2} - \frac{2}{4}\right) + \left(\frac{2}{3} - \frac{2}{5}\right) + \left(\frac{2}{4} - \frac{2}{6}\right) + \dots\right$
$\displaystyle \displaystyle\left + \left(\frac{2}{m-3} - \frac{2}{m-1}\right) + \left(\frac{2}{m-2} - \frac{2}{m}\right) + \left(\frac{2}{m-1} - \frac{2}{m+1}\right) + \left(\frac{2}{m} - \frac{2}{m + 2}\right)\right]$
$\displaystyle \displaystyle =\lim_{m \to \infty}\left(\frac{2}{1} + \frac{2}{2} - \frac{2}{m + 1} - \frac{2}{m + 2}\right)$
$\displaystyle \displaystyle = 3$.
Clearly this sum is convergent and converges to $\displaystyle \displaystyle 3$.
Hello, Reefer!
Quote:
$\displaystyle \displaystyle S \;=\;\sum ^{\infty}_{n = 1} \frac{4}{n^2+2n}$
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 \displaystyle S \;=\; 4\sum^{\infty}_{n=1}\frac{1}{n(n+2)} $
Partial Fraction Decomposition
We have: .$\displaystyle \dfrac{1}{n(n+2)} \:=\:\dfrac{A}{n} + \dfrac{B}{n+2}$
Multiply through by $\displaystyle n(n+2)\!:\;\;1 \;=\;A(n+2) + B(n) $
. . Let $\displaystyle n = 0\!:\;\;1 \:=\:A(0+2) + B(0) \quad\Rightarrow\quad A \:=\:\frac{1}{2}$
. . Let $\displaystyle n = -2:\;\;1 \:=\:A(0) + B(-2) \quad\Rightarrow\quad B \:=\:-\frac{1}{2}$
Hence: .$\displaystyle \displaystyle \frac{1}{n(n+2)} \;=\;\frac{\frac{1}{2}}{n} + \frac{-\frac{1}{2}}{n+2} \;=\;\frac{1}{2}\left(\frac{1}{n} - \frac{1}{n+2}\right)$
Then: .$\displaystyle \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 \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 $\displaystyle \frac{1}{1}$ and $\displaystyle \frac{1}{2}$
So we have: .$\displaystyle S \;=\;2\left(\frac{1}{1} + \frac{1}{2}\right) \;=\;2\left(\frac{3}{2}\right)$
. .Therefore: .$\displaystyle S \:=\:3$
The sum obviously doesn't converge to 0 since it is a sum of positive numbers!Quote:
Originally Posted by Reefer
$\displaystyle \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?
