Wikipedia:Reference desk/Archives/Mathematics/2016 June 20

= June 20 =

Harmonic series
Any known formula for $$\sum_{k=1}^n \frac{1}{k}$$ ? יהודה שמחה ולדמן (talk) 11:41, 20 June 2016 (UTC)
 * Harmonic number has a lot of information, is that what you are looking for? —Kusma (t·c) 12:07, 20 June 2016 (UTC)
 * The expression $$\sum_{k=1}^n\frac{1}{k}$$ is already a formula. If this doesn't count as a formula, then you haven't sufficiently clarified the rules for what a "formula" is.  I was also going to suggest that you look at harmonic number, but Kusma beat me to it.   Sławomir Biały  (talk) 12:21, 20 June 2016 (UTC)
 * While true, I find this somewhat pedantic. Usually when someone asks a question like this, the meaning is "a formula which allows easy calculation of the value". The original formula clearly cannot be easily computed for large n. -- Meni Rosenfeld (talk) 12:42, 20 June 2016 (UTC)
 * It's not clear that is what is intended. Is a closed form involving only some class of functions desired?  an asymptotic formula?  just any old formula?  The word "formula" conveys very little.  It is necessary to be pedantic to get clarity on what is desired.  You clearly interpreted the desired formula to be an asymptotic formula, but there are certainly others.   Sławomir Biały  (talk) 17:30, 20 June 2016 (UTC)
 * Not exactly. What the OP is really after (an exact expression in elementary form) does not exist AFAIK. So I settled for the next best thing, an elementary expression which asymptotically converges to the value. -- Meni Rosenfeld (talk) 17:43, 20 June 2016 (UTC)
 * A good approximation is $$\sum_{k=1}^n \frac1k \approx\log n+\gamma+\frac{1}{2n+1/3}$$, where gamma is the Euler–Mascheroni constant. In general it can be expanded to a Taylor series, which allows reaching high precision much faster than with computing the sum naively. -- Meni Rosenfeld (talk) 12:40, 20 June 2016 (UTC)