What is the sum of the following series:
SUMMATION(n=1 to infiniti) -1/(n^2 + 9n + 20)
I tried using the forumula:
sum = a/(1-r)
But i cant get an exact value for r so i am not sure
Thanks
$\displaystyle \frac{a}{1-r}$ is for a geometric series: $\displaystyle \sum_{n=1}^{\infty} ar^n$
To sum the series: $\displaystyle \sum_{n=1}^{\infty} \frac{-1}{(n+4)(n+5)}$, try working out the first few terms...
$\displaystyle \frac{-1}{30} + \frac{-1}{42} + \frac{-1}{56} + \frac{-1}{72} + \frac{-1}{90} + \frac{-1}{110} + \ldots$
Not sure what the value is, but it seems that it lies in $\displaystyle (-1,0)$
Hello
$\displaystyle n^2+9n+20=(n+4)(n+5)$
$\displaystyle \frac{-1}{(n+4)(n+5)}=\frac{n-n+4-5}{(n+4)(n+5)}=\frac{(n+4)-(n+5)}{(n+4)(n+5)}$
$\displaystyle =\frac{(n+4)}{(n+4)(n+5)}-\frac{(n+5)}{(n+4)(n+5)}=\frac{1}{n+5}-\frac{1}{n+4}$
So $\displaystyle \sum_{n=1}^\infty \frac{-1}{n^2+9n+20}=\sum_{n=1}^\infty \left(-\frac{1}{n+4}+\frac{1}{n+5}\right)$$\displaystyle =\left(-\frac 15+{\color{red}\frac 16}\right)+\left({\color{red}-\frac 16}+{\color{blue}\frac 17}\right)+\left({\color{blue}-\frac 17}+\frac 18\right)+\dots$
What can you conclude ? (this is called a telescoping sum)
@ colby : partial fractions is the key
I grouped the terms..
For the sum, you can notice that each term is annulated by the one which follows, apart from the first term, -1/5.
Telescoping series - Wikipedia, the free encyclopedia
$\displaystyle \frac{-1}{(n+4)(n+5)}=\frac{-1+5-5+n-n}{(n+4)(n+5)}=\frac{(n+4)-(n+5)}{(n+4)(n+5)}=$
$\displaystyle \frac{(n+4)}{(n+4)(n+5)}+\frac{-(n+5)}{(n+4)(n+5)} =\frac{1}{(n+5)}-\frac{1}{(n+4)}$
$\displaystyle \sum_{n=1}^{\infty} \left( \frac{1}{(n+5)} - \frac{1}{(n+4)} \right)= \sum_{n=1}^{\infty} \frac{1}{n+5} - \sum_{n=1}^{\infty} \frac{1}{n+4}$
lets focus on the 2nd sum for a second
$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n+4}=\frac{1}{5}+\sum_{n=2}^{\infty} \frac{1}{n+4}$
lets reindex the series k = n-1 theb n=k+1
$\displaystyle
\frac{1}{5}+\sum_{n=2}^{\infty} \frac{1}{n+4}=\frac{1}{5}+\sum_{k=1}^{\infty} \frac{1}{k+5}=\frac{1}{5}+\sum_{n=1}^{\infty} \frac{1}{n+5}
$
for the last step just reindex again with n=k
Now back to where we were before
$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n+5} - \sum_{n=1}^{\infty} \frac{1}{n+4}=\sum_{n=1}^{\infty} \frac{1}{n+5} -\left(\frac{1}{5}+ \sum_{n=1}^{\infty} \frac{1}{n+5} \right)$
$\displaystyle -\frac{1}{5}+\sum_{n=1}^{\infty}\left( \frac{1}{n+5}-\frac{1}{n-5}\right)=-\frac{1}{5}+\sum_{n=1}^{\infty}0=-\frac{1}{5}$