Sums of Series: Find Sum with n=1 to ∞

[lmannoia]

Homework Statement

2/(n+8)(n+6); find the sum of the series with n=1 to infinity.

Homework Equations

Hmm..

The Attempt at a Solution

I tried to split this into a partial fraction and try to find the limit, but for some reason that didn't work for me. I'm confused as to approach this, so any hints would be greatly appreciated. Thanks!

 

[Dick]

The reason why it does or doesn't work might be more clear if you show us what you got when you split it into a partial fraction.

 

[lmannoia]

Oh right, I definitely should have included that. I got 1/(n+6) - 1/(n+8) when I split it into a partial fraction. Now, I'm unsure of where to go from there.

 

[Dick]

Oh right, I definitely should have included that. I got 1/(n+6) - 1/(n+8) when I split it into a partial fraction. Now, I'm unsure of where to go from there.

Good. Now if you're still confused, write out the terms in 1/(n+6)-1/(n+8) out to say, n=5. Now start cancelling the repeated terms. It's a telescoping series, right? All but a finite number of terms just cancel.

 

[lmannoia]

I don't know why but I'm still confused. It seems like the terms are just randomly canceling out with no pattern. How is it that you come to an exact answer through telescoping?

 

[iamalexalright]

\frac{1}{7} - \frac{1}{8} + \frac{1}{8} - \frac{1}{9} + \frac{1}{9} - \frac{1}{10}...

Can you see the pattern?

 

[lmannoia]

I think so..everything is going to cancel but the first term and the final term?

 

[Dick]

I don't know why but I'm still confused. It seems like the terms are just randomly canceling out with no pattern. How is it that you come to an exact answer through telescoping?

They do cancel with a pattern. Take the n=4 terms. One is canceled by a term in the (4-2) term. The other is canceled by one in the (4+2) term. There is no contribution by terms with n>=4.

 

[Dick]

\frac{1}{7} - \frac{1}{8} + \frac{1}{8} - \frac{1}{9} + \frac{1}{9} - \frac{1}{10}...

Can you see the pattern?

That's a nice pattern. But it's not the right one for this series.

 

[iamalexalright]

Ahhhh, thought the second term was 1 / (1 + 7) - my mistak

 

[lmannoia]

They do cancel with a pattern. Take the n=4 terms. One is canceled by a term in the (4-2) term. The other is canceled by one in the (4+2) term. There is no contribution by terms with n>=4.

Wait, I don't understand what you mean when you say that there is no contribution by terms with n>=4. It's very possible that I'm doing it wrong, but when I started writing out the terms, to me it seems like they just keep canceling:
(1/7)-(1/9)+(1/8)-(1/10)+(1/9)-(1/11)+(1/10)-(1/12)+(1/11)-(1/13)+(1/12)-(1/14)+(1/3)...
Am I writing these out the wrong way? Or is there something that I just keep missing with this question?

 

[iamalexalright]

you are writing it correctly, let's see if you can see the pattern now

\frac{1}{7} - \frac{1}{9} + \frac{1}{8} - \frac{1}{10} + \frac{1}{9} - \frac{1}{11} + \frac{1}{10} - \frac{1}{12} + \frac{1}{11} - \frac{1}{13} + \frac{1}{12} - \frac{1}{14}if we simplify we see that we can get rid of -1/9 (n = 1) with +1/9 (n = 3), we cancel -1/10 (n = 2) with +1/10 (n = 4) and so on

can you now see what terms are left if we let n go to infinity?

 

[lmannoia]

GOT IT! Thank you both so much for your time and extreme patience!