# Sum of convergent series HELP

Sum of convergent series HELP!!

## Homework Statement

Find the sum of the convergent series -
the sum of 6 / (n+7)(n+9) from n=1 to infinity (∞)

A) 31/24
B) 45/56
C) 8/11
D) 17/24
E) 23/24

2. The attempt at a solution
I was looking in the book and they had one example that was kinda close so i tried to follow it and i did this....

= 6 times the sum of (1/(n+7))-(1/n+9))
= 6 times (1/8-1/10)+(1/9-1/11)... and so on

but that got me nowhere.

I tried plugging this into my calculator sum(seq(6/(x+7)(x+9),x,1,999,1)
and i got .7023779 which is the closest to answer D --> .70833333
but that's only up to term 999 and how do i know that the series won't get bigger than 17/24?

i found that up to the 7,490th term the sum is .7075417. which is closer so i'm going to go with answer D. :)

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## Answers and Replies

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Mark44
Mentor

Of the million ways you've tried, show us your work for one of them, and we'll give it our best.

HallsofIvy
Homework Helper

My thought would be to start with "partial fractions".
6/(x+9)(x+7)= A/(x+9)+ B/(x+7) so, multiplying by (x+9)(x+7), 6= A(x+7)+ B(x+9). If x= -7, 6= A(0)+ B(2) so B= 3. If x= -9, 6=A(-2)+ B(0) so A= -3:
6/(x+9)(x+7)= 3/(x+7)- 3/(x+9)= 3(1/(x+7)- 1/(x+9))
Sum 6/(x+9)(x+7)= 3[ sum(1/(x+7))- sum(1/(x+9)].
This gives a "telescoping series".

My thought would be to start with "partial fractions".
6/(x+9)(x+7)= A/(x+9)+ B/(x+7) so, multiplying by (x+9)(x+7), 6= A(x+7)+ B(x+9). If x= -7, 6= A(0)+ B(2) so B= 3. If x= -9, 6=A(-2)+ B(0) so A= -3:
6/(x+9)(x+7)= 3/(x+7)- 3/(x+9)= 3(1/(x+7)- 1/(x+9))
Sum 6/(x+9)(x+7)= 3[ sum(1/(x+7))- sum(1/(x+9)].
This gives a "telescoping series".

Telescoping series -
(b sub 1 - b sub 2)+(b sub 2 - b sub 3)+(b sub 3 - b sub 4)+ ...
Sum = (b sub 1) - lim (b sub n+1)

So with the telescoping series I get

3[ lim (1/8) - 1/(x+9)]

But that's not right.... umm....
are you sure that is a telescoping series?
(1/8 - 1/10)+(1/9 - 1/11)+(1/10 - 1/12)...
i thought the last # in the parenthesis has to be the same as the first # in the next parenthesis.

Mark44
Mentor

i thought the last # in the parenthesis has to be the same as the first # in the next parenthesis.
No, that doesn't have to be the case.
Your series in expanded form looks like this:
3[(1/8 - 1/10)+(1/9 - 1/11)+(1/10 - 1/12) + (1/11 - 1/13) + (1/12 - 1/14) + ... + (1/(n + 7) - 1/(n + 9) + ...]

You should notice that the only terms that are left after doing the subtractions are 1/8, 1/9, and -1/(n + 9), so the sequence of partial sums looks like this:
Sn = 3(1/8 + 1/9 - 1/(n + 9))
As n grows large without bound, what does Sn approach?

oh so it would be 3(17/72)= 17/24
i get it.. Thank you very much!