1. ## 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

2. 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 $(-1,0)$

3. Hello

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

4. 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?

5. Originally Posted by Moo
@ colby : partial fractions is the key
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 $-1 = n - n - 5 + 4$

6. 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

7. i cant conclude anything, all i see is a series of numbers with 1/7 repeated?

8. 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

9. but that doesnt give me the sum.

10. Therefore, there remains only -1/5 at the end
From Moo. This is the result of your infinite sum.

11. 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.

12. 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

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

$-\frac{1}{5}+\sum_{n=1}^{\infty}\left( \frac{1}{n+5}-\frac{1}{n-5}\right)=-\frac{1}{5}+\sum_{n=1}^{\infty}0=-\frac{1}{5}$

13. 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 $\sum \frac{{\color{red}-}1}{n^2+9n+20}$

14. oh sorry, this si totally my fault, typo
correct question is:

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

thanks

15. 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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