V venialove Junior Member Joined Mar 30, 2008 Messages 53 Mar 30, 2008 #1 Find the limit of s(n) as N-->? Let s(n) = _n_ \ (1+i/n)^2(1/n) /__ i=1
O o_O Full Member Joined Oct 20, 2007 Messages 393 Mar 30, 2008 #2 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: \(\displaystyle \sum_{i = 1}^{n} i = \frac{n(n+1)}{2}\) \(\displaystyle \sum_{i = 1}^{n} i^{2} = \frac{n(n+1)(2n+1)}{6}\) ---------------- Or if you recognize how definite integrals are represented as Riemann sums, you'll see that your sum is: \(\displaystyle \int_{0}^{1} (1 + x)^{2} dx\)
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: \(\displaystyle \sum_{i = 1}^{n} i = \frac{n(n+1)}{2}\) \(\displaystyle \sum_{i = 1}^{n} i^{2} = \frac{n(n+1)(2n+1)}{6}\) ---------------- Or if you recognize how definite integrals are represented as Riemann sums, you'll see that your sum is: \(\displaystyle \int_{0}^{1} (1 + x)^{2} dx\)
