TITLE: Is there any formula for the series $1 + \frac12 + \frac13 + \cdots + \frac 1 n = ?$
QUESTION [27 upvotes]: Is there any formula for this series?

$$1 + \frac12 + \frac13 + \cdots + \frac 1 n .$$

REPLY [11 votes]: Here is a way to interpret the harmonic numbers combinatorially:
$$H_n=\dfrac {\genfrac{[}{]}{0pt}{}{n+1}{2}}{n!}$$
where $\genfrac{[}{]}{0pt}{}{n+1}{2}$ is the absolute value of the Stirling number of the first kind, namely, the number of permutations of ${1,2,\dots,n,n+1}$ that have exactly $2$ cycles.
These satisfy the following recurrence:
$$\genfrac{[}{]}{0pt}{}{n}{k}=\genfrac{[}{]}{0pt}{}{n-1}{k-1}+(n-1)\genfrac{[}{]}{0pt}{}{n-1}{k}$$ which makes it algebraically obvious that they are related to the Harmonic numbers the way they are, though there is also a purely combinatorial proof. The virtue of this interpretation is that you can prove a whole host of crazy identities involving the Harmonic numbers by just translating the natural identities for the Stirling numbers.
The Stirling numbers of the first kind are also the coefficients of $x(x-1)(x-2)\cdots(x-(n-1))$, so their absolute values are the coefficients of $x(x+1)(x+2)\cdots (x+n-1)$, and so in particular

The nth Harmonic number is the coefficient of $x^2$ in $\frac1{n!}x(x+1)(x+2)\dots(x+n)$.

All of this can be found in Chapter 7 of Benjamin & Quinn's wonderful book Proofs That Really Count.