calculate-lim-n-k-1-2n-k-k-2-n-2-

Question Number 111754 by mathmax by abdo last updated on 04/Sep/20

calculate \lim_{n \to + \infty} \sum_{k = 1}^{2 n} \frac{k}{k^{2} + n^{2}}

Answered by Dwaipayan Shikari last updated on 04/Sep/20

\lim_{n \to \infty} \frac{1}{n} \sum^{2 n} \frac{k}{\frac{k^{2}}{n} + n} = \lim_{n \to \infty} \frac{1}{n} \underset{k = 1}{\sum^{2 n}} \frac{1}{\frac{k}{n} + \frac{n}{k}} = \int_{0}^{2} \frac{1}{x + \frac{1}{x}} d x

= \frac{1}{2} \int_{0}^{2} \frac{2 x}{x^{2} + 1} d x

= {[\frac{1}{2} l o g (x^{2} + 1)]}_{0}^{2}

= \frac{1}{2} l o g (5)

Answered by mathmax by abdo last updated on 07/Sep/20

A_{n} = \sum_{k = 1}^{n} \frac{k}{k^{2} + n^{2}} + \sum_{k = n + 1}^{2 n} \frac{k}{k^{2} + n^{2}} = S_{1}^{n} + S_{2}^{n}

S_{1}^{n} = \sum_{k = 1}^{n} \frac{k}{k^{2} (1 + \frac{n^{2}}{k^{2}})} = \frac{1}{n} \sum_{k = 1}^{n} \frac{n}{k (1 + \frac{1}{\frac{n^{2}}{k^{2}}})} \to \int_{0}^{1} \frac{x}{1 + \frac{1}{x}} dx

= \int_{0}^{1} \frac{x^{2}}{x + 1} dx = \int_{0}^{1} \frac{x^{2} - 1 + 1}{x + 1} dx = \int_{0}^{1} (x - 1) dx + \int_{0}^{1} \frac{dx}{x + 1}

= {[\frac{x^{2}}{2} - x]}_{0}^{1} + {[\ln (x + 1)]}_{0}^{1} = - \frac{1}{2} + \ln (2)

S_{2}^{n} = \sum_{k = n + 1}^{2 n} \frac{k}{k^{2} + n^{2}} =_{k - n = p} \sum_{p = 1}^{n} \frac{n + p}{{(n + p)}^{2} + n^{2}}

= \sum_{p = 1}^{n} \frac{n + p}{n^{2} + 2 np + p^{2} + n^{2}} = \sum_{p = 1}^{n} \frac{n + p}{{2 n}^{2} + 2 np + p^{2}}

= \sum_{p = 1}^{n} \frac{n (1 + \frac{p}{n})}{n^{2} (2 + \frac{2 p}{n} + \frac{p^{2}}{n^{2}})} = \frac{1}{n} \sum_{p = 1}^{n} \frac{1 + \frac{p}{n}}{{(\frac{p}{n})}^{2} + 2 (\frac{p}{n}) + 2} \to \int_{0}^{1} \frac{1 + x}{x^{2} + 2 x + 2} dx

= \frac{1}{2} \int_{0}^{1} \frac{2 x + 2}{x^{2} + 2 x + 2} dx = \frac{1}{2} {[\ln (x^{2} + 2 x + 2)]}_{0}^{1} = \frac{1}{2} (\ln (5) - \ln (2)) \Rightarrow

\lim_{n \to + \infty} A_{n} = \ln (2) - \frac{1}{2} + \frac{1}{2} \ln 5 - \frac{1}{2} \ln 2 = \frac{1}{2} \ln 2 + \frac{1}{2} \ln 5 - \frac{1}{2}

= \ln (\sqrt{10}) - \frac{1}{2}