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

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- May 21st 2008, 08:36 AMtaurusSum of series
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 - May 21st 2008, 08:43 AMcolby2152
$\displaystyle \frac{a}{1-r}$ 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 $\displaystyle (-1,0)$ - May 21st 2008, 08:43 AMMoo
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 :D - May 21st 2008, 09:03 AMtaurus
I still dont get it.

What happened to your first line with n-n etc?

Also i still dont get how to get the sum then? - May 21st 2008, 09:09 AMcolby2152
- May 21st 2008, 09:15 AMMoo
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 - May 21st 2008, 09:17 AMtaurus
i cant conclude anything, all i see is a series of numbers with 1/7 repeated?

- May 21st 2008, 09:23 AMMoo
- May 21st 2008, 09:30 AMtaurus
but that doesnt give me the sum.

- May 21st 2008, 09:39 AMarbolisQuote:

Therefore, there remains only -1/5 at the end

- May 21st 2008, 09:48 AMtaurus
But when i enter that as an answer it says its wrong it says:

Your answer is not correct.

Hint: break up the summand as partial fractions and then write out the series to the first few terms. - May 21st 2008, 09:52 AMTheEmptySet

$\displaystyle \frac{-1}{(n+4)(n+5)}=\frac{-1+5-5+n-n}{(n+4)(n+5)}=\frac{(n+4)-(n+5)}{(n+4)(n+5)}=$

$\displaystyle \frac{(n+4)}{(n+4)(n+5)}+\frac{-(n+5)}{(n+4)(n+5)} =\frac{1}{(n+5)}-\frac{1}{(n+4)}$

$\displaystyle \sum_{n=1}^{\infty} \left( \frac{1}{(n+5)} - \frac{1}{(n+4)} \right)= \sum_{n=1}^{\infty} \frac{1}{n+5} - \sum_{n=1}^{\infty} \frac{1}{n+4}$

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

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n+5} - \sum_{n=1}^{\infty} \frac{1}{n+4}=\sum_{n=1}^{\infty} \frac{1}{n+5} -\left(\frac{1}{5}+ \sum_{n=1}^{\infty} \frac{1}{n+5} \right)$

$\displaystyle -\frac{1}{5}+\sum_{n=1}^{\infty}\left( \frac{1}{n+5}-\frac{1}{n-5}\right)=-\frac{1}{5}+\sum_{n=1}^{\infty}0=-\frac{1}{5}$ - May 21st 2008, 09:54 AMMoo
- May 21st 2008, 09:57 AMtaurus
oh sorry, this si totally my fault, typo

correct question is:

SUMMATION(n=1 to infiniti) -2/(n^2 + 9n + 20)

?

thanks - May 21st 2008, 09:58 AMMoo