Question

Find the sum (n=1 to infinity) of 4/ (4n-3)(4n+1)
It's telescoping series. I understand how to do partial decomposition but I'm struggling to find the pattern

Answer 1

What did you get for your Partial Fraction Decomposition?

Answer 2

1/4n-3 - 1/4n+2

Answer 3

*4n+1 for the second

Answer 4

I sincerely hope that is not what you achieved for your partial fraction decomposition. Perhaps this? 1/(4n-3) - 1/(4n+1) The parentheses are NOT optional. Please remember your Order of Operations.

Okay, now write out the terms for the first three values of n.

Answer 5

n=1 1 - (1/5)
n=2 (1/5) - (1/9)
n=3 (1/9) - (1/13)

Answer 6

Do you see it? \(1 - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{9} + \dfrac{1}{9} - \dfrac{1}{13} = ...\)

Since we already know it converges - we do know that, right? - we can rearrange as we chose. Pick the first term, then pair each two after that. What happens?

Answer 7

Sn= 1 - (1/4n+1) right? so when you take the limit as n approaches infinity, it goes to 1. Which makes it convergent?

Answer 8

No, no, no and no. If we are going to rearrange it, we must ALREADY know it converges. It is actually finding the limit that we accomplish with the rearrangement. I guess we forgot to prove it converges, first?

Be that as it may, your very correct answer, \(S_{n} = 1 - \dfrac{1}{4n+1}\) demonstrates that the limit is 1 (one).

btw - You still put the parentheses in the wrong place. 1/4n + 1 = \(\dfrac{1}{4}\cdot n + 1\). Think long and hard about the Order of Operations. It is important at least for clarity of communication.

Answer 9

yeah, i forgot to prove that the limit converges first. What would be the best test to use for this one?

Answer 10

It's a nice, quadratic denominator, so it shouldn't be difficult to select an appropriate comparison.

Answer 11

Thank you!