AP Calc: Find lim[N->infty][sum[i=1,n][1+i/n)^2(1/n)]]

[venialove]

Find the limit of s(n) as N-->?

Let s(n) = _n_
\ (1+i/n)^2(1/n)
/__
i=1

 

[o_O]

Re: math ap calc

$\displaystyle \lim_{n \to \infty} \sum_{i = 1}^{n} \left(1 + \frac{i}{n}\right)^{2} \left(\frac{1}{n}\right)$

\(\displaystyle = \lim_{n \to \infty} \left(\frac{1}{n}\right) \sum_{i = 1}^{n} \left(1 + \frac{i}{n}\right)^{2} \quad \quad \frac{1}{n} \mbox{ acts as a constant in the summation}\)

$\displaystyle = \lim_{n \to \infty} \left(\frac{1}{n}\right) \sum_{i = 1}^{n} \left(\frac{i^{2}}{n^{2}} + \frac{2i}{n} + 1\right)$

$\displaystyle = \lim_{n \to \infty} \left(\frac{1}{n}\right) \left(\sum_{i = 1}^{n}\frac{i^{2}}{n^{2}} + \sum_{i = 1}^{n} \frac{2i}{n} + \sum_{i = 1}^{n}1\right)$

Pull out the "n" terms as they act as a constant in the summations and you should be able to simplify. I presume you already know these summations:

$\displaystyle \sum_{i = 1}^{n} i = \frac{n(n+1)}{2}$

$\displaystyle \sum_{i = 1}^{n} i^{2} = \frac{n(n+1)(2n+1)}{6}$

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Or if you recognize how definite integrals are represented as Riemann sums, you'll see that your sum is:

$\displaystyle \int_{0}^{1} (1 + x)^{2} dx$