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

Is there any formula for this series?

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

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–  Isaac Sep 20 '10 at 2:20
–  Ｊ. Ｍ. Sep 20 '10 at 2:29
One interesting thing in this series, and with all infinite series (I think), is that you can take away any finite number of members and the whole summation will still be equal to infinite. –  JMCF125 Jun 9 '13 at 16:56
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## 5 Answers

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.

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+1. There's no formula significantly different from 1+1/2+1/3+…+1/n, that is. :-) –  ShreevatsaR Sep 20 '10 at 3:21
Well, there's the digamma function, but it's still cheating I think. :) –  Ｊ. Ｍ. Sep 20 '10 at 3:32
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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...)

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Here's good information on computing harmonic numbers, either exactly or approximately. fredrik-j.blogspot.com/2009/02/… –  Charles Sep 21 '10 at 14:37
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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.

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This question made me recall this wonderful formula for $H_n$ due to Gregory Fontana:

$$H_n = \gamma + \log n + {1 \over 2n} - \sum_{k=2}^\infty { (k-1)! C_k \over n(n+1)\ldots(n+k-1)}, \qquad \textrm{ for } n=1,2,3,\ldots,$$

where the coefficients $C_k$ are the Gregory coefficients given by $${ z \over \log(1-z)} = \sum_{k=0}^\infty C_k z^k \qquad \textrm{ for } |z|<1.$$

Although I remember proving this as a student, about 30 years ago, I've just spent a frustrating 90 minutes trying to recover the proof without success, so unfortunately I cannot sketch it here. PS: I'll post a question shortly, and perhaps someone can beat me to it and put me out of my misery :-)

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Well there are some formulae for the harmonic numbers. For example, one of the most well known is $H_n = \int_0^1 {1 - x^n \over 1 - x} dx$, which has certain useful properties. However, it is not in any way simpler than the original definition. And if you're looking for numerical approximations rather than exact identities, you certainly are better off with the approximation $H_n= \ln (n+1/2) +\gamma + \epsilon (n)$

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