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I was reading about how the harmonic numbers are analogues to the logarithm in that $\displaystyle \log(x) = \int \frac{1}{x}dx$ and $\displaystyle H_x = \sum \frac{1}{1+x} \delta x$

Where indefinite summation is the inverse operator of $\Delta f(x) = f(x+1) - f(x)$ and $\frac{1}{x+1} = x^{\underline{-1}}$

The Wikipedia page derives the following Newton series:

$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \frac{x^{\underline{k}}}{k!} \hspace{10mm} (1)$$

But even though the proof makes sense I feel like it should have been $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} {x^{\underline{k}}} \hspace{10mm} (2)$$

As it mirrors the familiar Mercator series for the logarithm.

In other cases such as the "discrete exp" $2^x$ we have the Newton series which works nicely thanks to the binomial theorem.

$$\sum_{k=0}^{\infty} \frac{x^{\underline{k}}}{k!}$$ which perfectly mirrors $e^x$ 's power series.

My question is, if (1) is right series, what does (2) define, if anything, and why the discrepancy?

Bernard
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Theo Diamantakis
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    The summation must start with $k=1$ and according to Wolfram alpha, the result is : https://www.wolframalpha.com/input/?i=sum(k%3D1,infinity,(-1)%5E(k%2B1)*x%5Ek%2Fk!%2Fk) – Peter Jan 04 '18 at 16:34
  • thanks for spotting fixed now, your link is for something different though no? – Theo Diamantakis Jan 04 '18 at 16:36
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    Not sure,whether the link works properly : The function is $$f(x)=\ln(x)+\Gamma(0,x)+\gamma$$ where $\gamma$ is the Euler-Mascheroni-constant and $\Gamma(x,y)$ the incomplete Gamma-function – Peter Jan 04 '18 at 16:38
  • Very interesting, is can you suggest any resources on this or a proof? Any information is extremely sparse when talking about newton series for special functions and its usually a side note for some other application. – Theo Diamantakis Jan 04 '18 at 16:40
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    A proof will require the series of the incomplete gamma-function, apparently it must begin with $-\gamma-\ln(x)$ , containing the diverging (for $x\rightarrow 0$) term $\ln(x)$. We have $$\Gamma(0,x)=\int_x^{\infty} \frac{e^{-t}}{t} dt$$ Maybe this helps. – Peter Jan 04 '18 at 16:45
  • No problem, my curiosity is satisfied at least and your answer was helpful. Maybe someone else can also answer. – Theo Diamantakis Jan 04 '18 at 16:49
  • Did you notice my edit ? This should be the key to proof the desired series. – Peter Jan 04 '18 at 17:03
  • Yes, I can't say that I am an expert in manipulating the incomplete gamma function though. I was thinking about using the Euler Mclaurin formula to turn the integral into a series and write the logarithm as a power series, but the variable in the lower limit sort of makes things difficult. – Theo Diamantakis Jan 04 '18 at 17:08
  • Well, the derivate of $\Gamma(0,x)$ is just $\large \frac{e^{-x}}{x}$ which can be easily converted into a Laurent-series (the only non-polynomial term will be $\frac{1}{x}$), so the only problem will be to establish the constant $\gamma$ – Peter Jan 04 '18 at 17:17
  • $f(1) = \log(2)$ by the alternating harmonic series, and $\Gamma(0,1) = -Ei(-1)$. So $\log(2) = -Ei(-1) + \gamma$ supposedly, but according to Wolfram this is not so. – Theo Diamantakis Jan 04 '18 at 17:22
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    @Theo $f(1)=1$, observe that $x^\underline k=0$ if $x\in\Bbb N$ and $k>x$. –  Jan 04 '18 at 17:35
  • careless error thanks, was thinking in terms of normal powers. – Theo Diamantakis Jan 04 '18 at 17:36

2 Answers2

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Well, I didnt noticed before but observe that if $x\in\Bbb C\setminus\Bbb N$ then the sequence $\{(-1)^{k+1}x^\underline k/k\}_{k\in\Bbb N}$ doesnt converges in $\Bbb C$. In any case: $\lim_{n\to\infty}|x^\underline n/n|=\infty$ for any $x\in\Bbb C\setminus\Bbb N$, then these sequences cannot define a convergent series.

By the other hand, when $x\in\Bbb N$, then $x^\underline k=0$ when $k>x$, so the series reduces to a finite sum, that is

$$\sum_{k=1}^\infty(-1)^{k+1}\frac{x^\underline k}k=\sum_{k=1}^x(-1)^{k+1}\frac{x^\underline k}k,\quad\text{when }x\in\Bbb N$$

Then certainly your series in $(2)$ fails to define a function that represent the generalized harmonic numbers.

Also observe that, contrary to the Mercator series, your series defined by $(1)$ converges for a wide range of values of $x$ (I guess, but not totally sure now, for all complex $x$ such that ${\rm Re}(x)>0$). Thus it cannot be compared directly to the Mercator series who radius of convergence is just $1$.

  • Interesting, I guess the mercator series must be quite sensitive to perturbations so trying to replace the powers with falling powers breaks everything. – Theo Diamantakis Jan 04 '18 at 21:07
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As mentioned in the other response, there exist ways to write the function as a finite sum. As for the discrepancy, I want to add that while the sum may look similar to the Mercator series, once you adjust for the offset, assuming they're equal would imply that $\int\frac{1}{t+1}dt = \sum\frac{1}{n+1}$, which just isn't the case.

I would also suggest the following line of reasoning for the given Newton Series, stemming from the fact that $\sum_{n=0}^\infty\frac{a^n}{n!}x^{\underline{n}} = (1+a)^x$ for $|a|<1$, which at the boundary $a=-1$ converges to 0 for all positive values of x. $$\sum_{n=0}^\infty-\frac{(-1)^n}{n!}x^{\underline{n}}=-\frac{1}{0!}x^{\underline{0}}+\frac{1}{1!}x^{\underline{1}}-\frac{1}{2!}x^{\underline{2}}+...=0$$ $$\sum_{n=1}^\infty-\frac{(-1)^n}{n!}x^{\underline{n}}=\frac{1}{1!}x^{\underline{1}}-\frac{1}{2!}x^{\underline{2}}+\frac{1}{3!}x^{\underline{3}}-...=1$$ $$=\frac{1}{1!}x-\frac{1}{2!}x(x-1)+\frac{1}{3!}x(x-1)(x-2)-...=1$$ $$\frac{1}{1!}-\frac{1}{2!}(x-1)+\frac{1}{3!}(x-1)(x-2)-...=\frac{1}{x}$$ $$\sum_{n=0}^\infty\frac{(-1)^n}{(n+1)!}x^{\underline{n}}=\frac{1}{1!}-\frac{1}{2!}x+\frac{1}{3!}x(x-1)-...=\frac{1}{x+1}$$ $$\sum_{n=1}^\infty-\frac{(-1)^n}{n!\cdot{n}}x^{\underline{n}}=\frac{1}{1!\cdot{1}}x^{\underline{1}}-\frac{1}{2!\cdot{2}}x^{\underline{2}}+\frac{1}{3!\cdot{3}}x^{\underline{3}}-...=\sum\frac{1}{x+1}$$

GlumTang
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