[Analiza] oblicz granice
: 29 paź 2007, o 17:40
obliczyć \(\displaystyle{ \lim_{n \to } n(n(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}-\ln2)-0,25)}\)
Skorzystajmy z:
1. \(\displaystyle{ \frac{1}{n+1} + \ldots + \frac{1}{2n} = H_{2n} - H_n}\)
2. \(\displaystyle{ H_n = \gamma + \ln n + \frac{1}{2n} + \mathcal{O} \left( \frac{1}{n^2} \right)}\)
gdzie \(\displaystyle{ H_n}\) to n-ta liczba harmoniczna.
Granicę możemy zatem zapisać jako:
\(\displaystyle{ \lim_{n \to + } \left\{ n^2 \left( H_{2n} - H_n - \ln 2 \right) - \frac{n}{4} \right\} \; =\\
\lim_{n \to + } \left\{ n^2 \left( \gamma + \ln 2n + \frac{1}{4n} + \mathcal{O} \left( \frac{1}{n^2} \right) - \gamma - \ln n - \frac{1}{2n} + \mathcal{O} \left( \frac{1}{n^2} \right) - \ln 2 \right) - \frac{n}{4} \right\} =\\
\lim_{n \to + } \left\{ n^2 \left( - \frac{1}{4 n} + \mathcal{O} \left( \frac{1}{n^2} \right) \right) - \frac{n}{4} \right\} \; = \\
\lim_{n \to + } \left( - \frac{n}{2} + \mathcal{O} (1) \right) \; = \; - }\)