# Thread: Proving Summation is equal to expression for all positive integers n

1. ## Proving Summation is equal to expression for all positive integers n

Prove that ∑ from k =1 to n of 1/(4k^2 - 1) = n/(2n+1) for all positive integers n.

In words: Prove that the sum (from k =1 to n) of one over (4 times k squared minus one) is equal to n over two times n + 1

I am beyond beyond beyond lost.

2. ## Re: Proving Summation is equal to expression for all positive integers n

Originally Posted by azollner95
Prove that ∑ from k =1 to n of 1/(4k^2 - 1) = n/(2n+1) for all positive integers n.
$\dfrac{1}{4x^2-1}=\dfrac{2^{-1}}{2k-1}-\dfrac{2^{-1}}{2k+1}$
Then look at collapsing sums.

3. ## Re: Proving Summation is equal to expression for all positive integers n

What is a collapsing sum and how do you get there. Lost on a few of the basic steps needed. Is it similar to telescoping series?

4. ## Re: Proving Summation is equal to expression for all positive integers n

Originally Posted by azollner95
a collapsing sum = to telescoping series?
Samething

$\frac{1}{2}\left[ {\left( {\frac{1}{1} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{5}} \right) + \left( {\frac{1}{5} - \frac{1}{7}} \right) \cdots + \left( {\frac{1}{{2n - 1}} - \frac{1}{{2n + 1}}} \right)} \right] = \frac{1}{2}\left[ {1 - \frac{1}{{2n + 1}}} \right]$

5. ## Re: Proving Summation is equal to expression for all positive integers n

Originally Posted by azollner95
I am beyond beyond beyond lost.
Are you saying your teacher gave you this problem
without any teaching?
Get him/her fired!