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(Fix math syntax error: change > to \geq in the inequality (non-strict, since 1/(2^m+2^m) = 1/2^{m+1}) (via update-page on MediaWiki MCP Server))
 
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While it does not occur often in classical mathematics, it crops up more often in algorithm analysis.
While it does not occur often in classical mathematics, it crops up more often in algorithm analysis.


We can make <math>H_n</math> as large as we please from observing that


<math>
H_{2^m} \geq 1 + \dfrac{m}{2}
</math>


This results from the fact that
<math>
H_{2^{m+1}} = H_{2^m} + \dfrac{1}{2^m+1} + \dfrac{1}{2^m + 2} + \dots + \dfrac{1}{2^{m+1}}
</math>
Now we have,
<math>
H_{2^m} + \dfrac{1}{2^m+1} + \dfrac{1}{2^m + 2} + \dots + \dfrac{1}{2^{m+1}} \geq H_{2^m} + \dfrac{1}{2^{m+1}} + \dfrac{1}{2^{m + 1}} + \dots + \dfrac{1}{2^{m+1}}
</math>
and the right side is
<math>
H_{2^m} + \frac{1}{2}
</math>
therefore
<math>
H_{2^m} \geq 1 + \dfrac{m}{2}
</math>


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Latest revision as of 01:22, 6 June 2026

Volume 1

Chapter 1: Basic Concepts: Harmonic numbers

Harmonic numbers become important in analyses of algorithms. Define

$ H_n = 1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dots + \dfrac{1}{n} = \sum_{1 \leq k \leq n} \dfrac{1}{k} \qquad n \geq 0 $

While it does not occur often in classical mathematics, it crops up more often in algorithm analysis.

We can make $ H_n $ as large as we please from observing that

$ H_{2^m} \geq 1 + \dfrac{m}{2} $

This results from the fact that

$ H_{2^{m+1}} = H_{2^m} + \dfrac{1}{2^m+1} + \dfrac{1}{2^m + 2} + \dots + \dfrac{1}{2^{m+1}} $

Now we have,

$ H_{2^m} + \dfrac{1}{2^m+1} + \dfrac{1}{2^m + 2} + \dots + \dfrac{1}{2^{m+1}} \geq H_{2^m} + \dfrac{1}{2^{m+1}} + \dfrac{1}{2^{m + 1}} + \dots + \dfrac{1}{2^{m+1}} $

and the right side is

$ H_{2^m} + \frac{1}{2} $

therefore

$ H_{2^m} \geq 1 + \dfrac{m}{2} $

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