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 The harmonic sequence
11-09-2010, 11:44 AM
Post: #1
 elim Moderator Posts: 581 Joined: Feb 2010 Reputation: 0
The harmonic sequence
Consider the sequence $H_{n}=1+\frac{1}{2}+\ldots +\frac{1}{n}, n\geq 1$. Calculate the limits:

i) $\lim_{n\to\infty}(\sqrt[n]{H_{n}})^{H_{n}}$;

ii) $\lim_{n\to\infty}\frac{n}{\ln^{2} n}\left[(\sqrt[n]{H_{n}})^{H_{n}}-1\right]$.

http://www.artofproblemsolving.com/Forum...&t=376710&
12-14-2010, 10:31 AM
Post: #2
 elim Moderator Posts: 581 Joined: Feb 2010 Reputation: 0
RE: The harmonic sequence
$\displaystyle{\lim_{n\to\infty}\frac{H_n}{\ln n}=\lim_{n\to\infty}\frac{\sum_{k=1}^n\frac{1}{k}}{\ln n}\;\overset{stolz}{=} \;1}$

$\displaystyle{\lim_{n\to\infty}\ln \left(\sqrt[n]{H_n} \right )^{H_n}=\lim_{n\to\infty}\frac{H_n \ln H_n}{n}=\lim_{n\to\infty}\frac{\ln n}{n}\left(\ln\frac{H_n}{\ln H_n}+\ln(\ln n) \right )=\lim_{n\to\infty}\frac{\ln n \ln(\ln n)}{n}=0}$

(1) $\displaystyle{\lim_{n\to\infty}\left(\sqrt[n]{H_n} \right )^{H_n}=1}$

$e^h -1 = e^{\theta h} = O(h)$
$\displaystyle{\left(\sqrt[n]{H_n} \right )^{H_n}-1 = O\left(\frac{(\ln n)\ln(\ln n)}{n} \right )}$

(2) $\displaystyle{0\le \lim_{n\to\infty}\frac{n}{\ln^2 n} \left(\left(\sqrt[n]{H_n} \right ) ^{H_n}-1\right )\le \lim_{n\to\infty} \frac{(\ln n)\ln(\ln n)}{\ln^2 n}}=0$
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