Sum of convergent series HELP!

[gitty_678]

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. :)

 

[Mark44]

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

 

[HallsofIvy]

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".

 

[gitty_678]

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]

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:
S_n = 3(1/8 + 1/9 - 1/(n + 9))
As n grows large without bound, what does S_n approach?

 

[gitty_678]

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