Check my Limit

**veronicak5678** · August 30th 2011, 05:59 PM

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.

**Sudharaka** · August 30th 2011, 06:02 PM

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

**veronicak5678** · August 30th 2011, 06:45 PM

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

**Prove It** · August 30th 2011, 07:48 PM

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

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

**veronicak5678** · August 30th 2011, 08:35 PM

Got it. Thanks very much.

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