Is there any formula for the series $1 + \frac12 + \frac13 + \cdots + \frac 1 n = ?$

Is there any formula for this series?

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

Solution 1:

There is no formula for the nth partial sum of the harmonic series, only approximations. A well known approximation by Euler is that the nth partial sum is approximately $$\ln(n) + \gamma $$ where $\gamma$ is the Euler–Mascheroni constant and is close to $0.5772$. The amount of error in this approximation gets arbitrarily small for sufficiently large values of $n$.

A well known fact in mathematics is that the harmonic series is divergent which means that if you add up enough terms in the series you can make their sum as large as you wish.

Solution 2:

Let $H_n=1+1/2+\cdots+1/n$, as usual. Then, for $n>1$,

$$H_n =\log(n+1/2) + \gamma + \varepsilon(n)$$

with

$$0 < \varepsilon(n) < 1/(24n^2)$$

so if you can use a high-precision estimate rather than an exact answer this may be good enough. For example, $1+1/2+1/3+\cdots+1/10^{100}$ is between

$$230.83572496430610126240565755851882319115230819881722120213855733164212869451291269453666757225157658376140985147843194582191305052276721850285291090752309248454422116629840542211766342541591511108644544$$

and

$$230.83572496430610126240565755851882319115230819881722120213855733164212869451291269453666757225157658376140985147843194582191305052276721850285291090752309248454422116629840542211766342541591511108644546$$

where the numbers are identical up to their last digit. If you need higher precision, a series expansion will yield formulas that are increasingly precise for large values of $n$. (For small values of $n$, calculate directly...)