Thread: Check my Limit

Please tell me if this is correct:

Find the limit of (E k=1 to infinity) of 1/ (4n^2 -1)

I believe this is 0.

Originally Posted by veronicak5678

Please tell me if this is correct:

Find the limit of (E k=1 to infinity) of 1/ (4n^2 -1)

I believe this is 0.

Hi veronicak5678,

Do you mean, $\displaystyle \sum_{k=1}^{\infty}\frac{1}{4n^2-1}$ ?

Yes, that's what I mean. Sorry it's not clear.

$\displaystyle \displaystyle \frac{1}{4n^2 - 1} = \frac{1}{(2n - 1)(2n+1)} = \frac{1}{2}\left(\frac{1}{2n - 1} - \frac{1}{2n + 1}\right)$, so

$\displaystyle \displaystyle \begin{align*} \sum_{n = 1}^{\infty}\frac{1}{4n^2 - 1} &= \frac{1}{2}\lim_{k \to \infty}\sum_{n = 1}^{k}\left(\frac{1}{2n-1} - \frac{1}{2n + 1}\right) \\ &= \frac{1}{2}\lim_{k \to \infty}\left[\left(1 - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{7}\right) + \dots + \left(\frac{1}{2k - 1} - \frac{1}{2k + 1}\right)\right] \\ &= \frac{1}{2}\lim_{k \to \infty}\left(1 - \frac{1}{2k + 1}\right) \\ &= \frac{1}{2}\left(1 - 0\right) \\ &= \frac{1}{2} \end{align*}$

Got it. Thanks very much.