Exact value seems to be hard!

Is there any way to find the exact value of \(A_n = 1 + \frac{1}{2} + \frac{1}{3} +...+ \frac{1}{n}\)? And is there a closed form?

Note by Steven Jim
1 year, 1 month ago

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This is called the Harmonic series, and there is no closed form. However, the sum \(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}\) is approximately \(\log n\). See the following for more details.

https://en.wikipedia.org/wiki/Harmonicseries(mathematics)

http://mathworld.wolfram.com/HarmonicSeries.html

Jon Haussmann - 1 year, 1 month ago

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Thanks! Very much appreciated!

Steven Jim - 1 year, 1 month ago

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