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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 3 months, 1 week 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 · 3 months, 1 week ago

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@Jon Haussmann – Thanks! Very much appreciated! – Steven Jim · 3 months, 1 week ago

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## Comments

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TopNewestThis 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/Harmonic

series(mathematics)http://mathworld.wolfram.com/HarmonicSeries.html – Jon Haussmann · 3 months, 1 week ago

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– Steven Jim · 3 months, 1 week ago

Thanks! Very much appreciated!Log in to reply