New user? Sign up

Existing user? Sign in

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 5 months, 2 weeks ago

Sort by:

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

Log in to reply

Thanks! Very much appreciated!

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

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

Log in to reply

Thanks! Very much appreciated!

Log in to reply