# Sum of series

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• May 21st 2008, 08:36 AM
taurus
Sum 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 AM
colby2152
Quote:

Originally Posted by taurus
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

$\frac{a}{1-r}$ is for a geometric series: $\sum_{n=1}^{\infty} ar^n$

To sum the series: $\sum_{n=1}^{\infty} \frac{-1}{(n+4)(n+5)}$, try working out the first few terms...

$\frac{-1}{30} + \frac{-1}{42} + \frac{-1}{56} + \frac{-1}{72} + \frac{-1}{90} + \frac{-1}{110} + \ldots$

Not sure what the value is, but it seems that it lies in $(-1,0)$
• May 21st 2008, 08:43 AM
Moo
Hello :)

Quote:

Originally Posted by taurus
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

$n^2+9n+20=(n+4)(n+5)$

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

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

So $\sum_{n=1}^\infty \frac{-1}{n^2+9n+20}=\sum_{n=1}^\infty \left(-\frac{1}{n+4}+\frac{1}{n+5}\right)$ $=\left(-\frac 15+{\color{red}\frac 16}\right)+\left({\color{red}-\frac 16}+{\color{blue}\frac 17}\right)+\left({\color{blue}-\frac 17}+\frac 18\right)+\dots$

What can you conclude ? (this is called a telescoping sum)

@ colby : partial fractions is the key :D
• May 21st 2008, 09:03 AM
taurus
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 AM
colby2152
Quote:

Originally Posted by Moo
@ colby : partial fractions is the key :D

I had to rush out of the room before I finished the solution. Thanks for finishing it for me.

Taurus, Moo did some algebraic manipulation by having $-1 = n - n - 5 + 4$
• May 21st 2008, 09:15 AM
Moo
Quote:

Originally Posted by taurus
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?

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 AM
taurus
i cant conclude anything, all i see is a series of numbers with 1/7 repeated?
• May 21st 2008, 09:23 AM
Moo
Quote:

Originally Posted by taurus
i cant conclude anything, all i see is a series of numbers with 1/7 repeated?

..

It's -1/5+1/6-1/6+1/7-1/8+1/8-1/9+1/9-..............

So you can see that fractions annulate themselves, apart from -1/5.

Therefore, there remains only -1/5 at the end
• May 21st 2008, 09:30 AM
taurus
but that doesnt give me the sum.
• May 21st 2008, 09:39 AM
arbolis
Quote:

Therefore, there remains only -1/5 at the end
From Moo. This is the result of your infinite sum.
• May 21st 2008, 09:48 AM
taurus
But when i enter that as an answer it says its wrong it says:
Hint: break up the summand as partial fractions and then write out the series to the first few terms.
• May 21st 2008, 09:52 AM
TheEmptySet
Quote:

Originally Posted by taurus
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

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

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

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

lets focus on the 2nd sum for a second

$\sum_{n=1}^{\infty} \frac{1}{n+4}=\frac{1}{5}+\sum_{n=2}^{\infty} \frac{1}{n+4}$

lets reindex the series k = n-1 theb n=k+1

$
\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

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

$-\frac{1}{5}+\sum_{n=1}^{\infty}\left( \frac{1}{n+5}-\frac{1}{n-5}\right)=-\frac{1}{5}+\sum_{n=1}^{\infty}0=-\frac{1}{5}$
• May 21st 2008, 09:54 AM
Moo
Quote:

Originally Posted by taurus
But when i enter that as an answer it says its wrong it says:
Hint: break up the summand as partial fractions and then write out the series to the first few terms.

This is what I did : break the summand...Then I wrote out the first few terms of the series...

I see no mistake ~

Was that $\sum \frac{1}{n^2+9n+20}$ or $\sum \frac{{\color{red}-}1}{n^2+9n+20}$
• May 21st 2008, 09:57 AM
taurus
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 AM
Moo
Quote:

Originally Posted by taurus
oh sorry, this si totally my fault, typo
correct question is:

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

thanks

$\sum \frac{-2}{n^2+9n+20}=2 \cdot \sum \frac{-1}{n^2+9n+20}=2 \cdot \left(-\frac 15\right)$

...
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