H-n-1-1-2-1-3-1-n-H-2n-compute-H-2n-H-n-and-H-n-1-H-n-

Question Number 180828 by Vynho last updated on 17/Nov/22

H_{n} = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}

H_{2 n} = ? c o m p u t e H_{2 n} - H_{n} a n d H_{n + 1} - H_{n}

Commented by Frix last updated on 17/Nov/22

obviously H_{n + 1} - H_{n} = \frac{1}{n + 1}

I believe that \lim_{n \to \infty} (H_{k n} - H_{n}) = \ln k for k \in N

Answered by Frix last updated on 19/Nov/22

H_{2 n} - H_{n} = \underset{k = 1}{\sum^{2 n}} \frac{1}{k} - \underset{k = 1}{\sum^{n}} \frac{1}{k} =

= \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{2 n} - \underset{k = 1}{\sum^{n}} \frac{1}{k} =

= \frac{1}{1} + \frac{1}{3} + \dots + \frac{1}{2 n - 1} + \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2 n} - \underset{k = 1}{\sum^{n}} \frac{1}{k} =

= \underset{k = 1}{\sum^{n}} \frac{1}{2 k - 1} + \underset{k = 1}{\sum^{n}} \frac{1}{2 k} - \underset{k = 1}{\sum^{n}} \frac{1}{k} =

= \underset{k = 1}{\sum^{n}} \frac{1}{2 k - 1} + \frac{1}{2} \underset{k = 1}{\sum^{n}} \frac{1}{k} - \underset{k = 1}{\sum^{n}} \frac{1}{k} =

= \underset{k = 1}{\sum^{n}} \frac{1}{2 k - 1} - \frac{1}{2} \underset{k = 1}{\sum^{n}} \frac{1}{k} = \underset{k = 1}{\sum^{n}} \frac{1}{2 k - 1} - \frac{H_{n}}{2}

\Rightarrow

H_{2 n} = \frac{H_{n}}{2} + \underset{k = 1}{\sum^{n}} \frac{1}{2 k - 1}

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