# Analytic continuation of harmonic series

Is there an accepted analytic continuation of $$\sum_{n=1}^m \frac{1}{n}$$? Even a continuation to positive reals would be of interested, though negative and complex arguments would also be interesting.

I don't have a specific application in mind, but I'd very much like to understand how / if such a continuation could be accomplished. I've Googled but haven't come up with anything meaningful - perhaps because it's not possible?

ADDENDUM

@Noble below suggests $$\frac{\Gamma'(x)}{\Gamma(x)}$$. But this produces the following mismatched plots: Can anyone explain?

• As is the case with the Gamma function, it would help to specify what properties you want the analytic continuation to retain. For instance, Euler tells us that $H_n=\int_0^1 \frac {1-x^n}{1-x}dx$. Replacing $n$ by a continuous parameter gives you an analytic continuation (at least for $n>0$). Is it useful? – lulu Jan 1 at 14:04
• @Richard Burke-Ward Here's the explanation: the correct comparison is: Plot[{EulerGamma + Gamma'[z + 1]/Gamma[z + 1], HarmonicNumber[z]}, {z, -1, 5}] – Dr. Wolfgang Hintze Jan 2 at 9:11

## 2 Answers

I am not sure if this is what you meant, but Wolfram Alpha has an analytic formula for the $$n^{\text{th}}$$ harmonic number: Here, the digamma function is $$\psi_0(x)=\frac{\Gamma'(x)}{\Gamma(x)}$$, which I believe is defined for all numbers in the complex plane except for negative real integers.

• Hi @Noble. The two don't seem to match when plotted - see my addendum to the OP. Many thanks though! – Richard Burke-Ward Jan 1 at 21:04
• @RichardBurke-Ward I am sorry, but you plotted the wrong function. You plotted just $\phi_0(n)$ vs $H_n$. However, what I proposed was $\gamma+\phi_0(n+1)$ vs $H_n$. Here is the correct plot on Wolfram Alpha: wolframalpha.com/input/… As you can see, the two plots coincide. – Noble Mushtak Jan 1 at 21:48
• Understood. Appreciated, both of you. – Richard Burke-Ward Jan 2 at 13:39

Let's try it in an elementary manner

1. We can use the defining recursion of the harmonic number valid for $$n\in Z^{+}$$

$$H_{n} = H_{n-1} + \frac{1}{n}, H_{1}=1\tag{1a}$$

also for any complex $$z$$

$$H_{z} = H_{z-1} + \frac{1}{z}, H_{1}=1\tag{1b}$$

For instance for $$z=1$$ we obtain $$H_{1} = H_{0} + \frac{1}{1}$$

from which we conclude that $$H_{0}=0$$.

If we try to find $$H_{-1}$$ we encounter the problem that from $$H_0 = 0 = \lim_{z\to0}(H_{-1+z} + \frac{1}{z})$$ we find that $$H_{z} \simeq \frac{1}{z}$$ for $$z\simeq 0$$. In other words, $$H_{z}$$ has a simple pole at $$z=-1$$.

Hence we cannot continue in this manner to go to further into the region of negative $$z$$, so let us move to the following general approach.

1. Starting with this formula for the harmonic number which is valid for $$n\in Z^{+}$$

$$H_{n} = \frac{1}{2}+ ... + \frac{1}{n}\\\\=\frac{1}{1}+ \frac{1}{2}+ ... + \frac{1}{n} +\frac{1}{1+n}+\frac{1}{n+2} + ... \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;-\frac{1}{1+n}- \frac{1}{n+2} + ...\\=\sum_{k=0}^\infty \left(\frac{1}{k}-\frac{1}{ (k+n)}\right)\tag{2}$$

The sum can be written as

$$H_{n}= \sum_{k=1}^\infty \frac{n}{k (k+n)}\tag{3}$$

and this can be extended immediately to complex values $$z$$ in place of $$n$$

$$H_{z}= \sum_{k=1}^\infty \frac{z}{k (k+z)}=\sum_{k=1}^\infty \left(\frac{1}{k} -\frac{1}{k+z}\right)\tag{4}$$

This sum is convergent (the proof is left to the reader) for any $$z$$ except for $$z=-1, -2, ...$$ where $$H_{z}$$ has simple poles with residue $$-1$$.

Hence $$(4)$$ gives the analytic continuation.

For instance close to $$z=0$$ we have as in 1. that

$$H_{z} \simeq z \sum_{k=1}^\infty \frac{1}{k^2} = z\;\zeta(2) =z\;\frac{\pi^2}{6}\to 0$$

We can also derive an integral representation from the second form of $$(4)$$ writing

$$\frac{1}{k} -\frac{1}{k+z} =\int_0^1 (x^{k-1}-x^{z+k-1})\,dx$$

Performing the sum under the integral is just doing a geometric sum and gives

$$H_{z} = \int_0^1 \frac{1-x^{z}}{1-x}\,dx \tag{5}$$

1. $$H_{z}$$ at negativ half integers ($$z = -\frac{1}{2}, -\frac{3}{2}, ...$$)

These can be calculated from $$(1b)$$ as soon as $$H_{\frac{1}{2}}$$ is known.

So let us calculate $$H_\frac{1}{2}$$.

Consider

$$H_{2n} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{2n}$$

Splitting even and odd terms gives

$$H_{2n}= \frac{1}{1} + \frac{1}{3} + \frac{1}{5} + ... + \frac{1}{2n-1}\\+ \frac{1}{2} + \frac{1}{4} + ... + \frac{1}{2n}\\= \sum_{k=1}^n \frac{1}{2k-1} + \frac{1}{2} H_{n}\tag{6}$$

Now for the sum of the odd terms we write as in $$(1)$$

$$O_{n} = \sum_{k=1}^\infty \left(\frac{1}{2k-1} - \frac{1}{2(n+k)-1}\right)\tag{7}$$

This can be anlytically continued to any complex $$n\to z$$.

Replacing as before the summand by an integral and doing the summation under the integral gives

$$O_{z} = \int_0^1 \frac{1-x^{2z}}{1-x^2}\,dx\tag{8}$$

Substituting $$x \to \sqrt{t}$$ we find

$$O_{z} = \frac{1}{2}\int_0^1 \frac{1}{\sqrt{t}}\frac{1-t^{z}}{1-t}\,dt\\= \frac{1}{2}\int_0^1 \frac{1-t^{z-\frac{1}{2}}}{1-t}\,dt- \frac{1}{2}\int_0^1 \frac{1-t^{-\frac{1}{2}}}{1-t}\,dt\\ =\frac{1}{2}H_{z-\frac{1}{2}}+\log{2}\tag{9}$$

Hence $$(6)$$ can be written as

$$H_{2z} = \frac{1}{2} H_{z} +\frac{1}{2} H_{z-\frac{1}{2}}+\log{2}\tag{10}$$

Letting $$z=1$$ this gives

$$H_{2} = \frac{1}{2} H_{1} +\frac{1}{2} H_{\frac{1}{2}}+\log{2}$$

from which we deduce finally

$$H_{\frac{1}{2}} = 2(1-\log{2})\simeq 0.613706 \tag{11}$$

EDIT

Altenatively, the calculation of $$H_{\frac{1}{2}}$$ can be done using $$(5)$$ with the substitution $$(x\to t^2)$$:

$$H_{\frac{1}{2}} = \int_0^1 \frac{1-x^{\frac{1}{2}}}{1-x}\,dx = 2\int_0^1 t \frac{(1-t)}{{1-t^2}}\,dt = 2\int_0^1 t \frac{(1-t)}{(1+t)(1-t)}\,dt \\=2\int_0^1 \frac{t}{{1+t}}\,dt=2\int_0^1 \frac{1+t}{{1+t}}\,dt -2\int_0^1 \frac{1}{{1+t}}\,dt = 2 - 2 \log(2)$$

and we have recovered $$(11)$$.

As an exercise calculate $$H_{\frac{1}{n}}$$ for $$n =3, 4,...$$.

I found that Mathematica returns explicit expression up to $$n=12$$ except for the case $$n=5$$. I have not yet unerstood the reason for this exception. Maybe someone else can explain it?

• I think there's a typo in (2) and similar, the sums should start from 1, not 0. – Bladewood Jan 2 at 1:32
• @Bladewood You are right. Thanks. Have corrected it. – Dr. Wolfgang Hintze Jan 2 at 8:46