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Lim z ln(2)
: 23 lis 2025, o 12:40
autor: mol_ksiazkowy
Udowodnić, że \(\displaystyle{ \lim_{n \to +\infty} ( \frac{2^{ \frac{1}{n} }}{n+1} + \frac{2^{ \frac{2}{n} }}{n+ \frac{1}{2} } +...+ \frac{2^{ \frac{n}{n} }}{n+\frac{1}{n}} ) = \frac{1}{\ln(2)}}\) .
Re: Lim z ln(2)
: 24 lis 2025, o 11:14
autor: trol-24-11-2025-2
\(\displaystyle{ \frac{n}{n+1} \cdot \frac{1}{n} \sum_{k=1}^{n} 2^{ \frac{k}{n} } \le \sum_{k=1}^{n} \frac{2^{ \frac{k}{n} }}{n+ \frac{1}{k} } \le \frac{1}{n} \sum_{k=1}^{n} 2^{ \frac{k}{n} } }\)
\(\displaystyle{ \frac{1}{n} \sum_{k=1}^{n} 2^{ \frac{k}{n} } \rightarrow \int_{0}^{1} 2^xdx= \frac{2^x}{\ln 2} |^1_{0}= \frac{2}{\ln 2} - \frac{1}{\ln 2} = \frac{1}{\ln 2} }\)
to dąży do całki Riemanna...
\(\displaystyle{ \frac{n}{n+1} \rightarrow 1}\)
więc:
\(\displaystyle{ \sum_{k=1}^{n} \frac{2^{ \frac{k}{n} }}{n+ \frac{1}{k} } \rightarrow \frac{1}{\ln 2} }\)