Convergence of Digamma Functions.

Today I would like to share another amazing property (convergence) of Digamma Functions. I found this one while playing with some integrals and their series expansion :

For all x>0x > 0

limnψ(nx+n+1)ψ(nx+1)=log(1+1x)\lim_{n \rightarrow \infty} \psi(nx+n+1)-\psi(nx+1) = \log \left(1+\frac{1}{x}\right)

Proof.

Great work by my friends, Ronak and Pratik.

So, here' my method.

Proof 1

We'll compute the following sum in two different ways limnr=1n1nx+r\lim_{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{nx+r}

We can convert the above riemann sum into integral as :

limnr=1n1nx+r=01dtx+t=log(x+t)01=log(1+1x)\lim_{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{nx+r} = \int_0^1 \frac{dt}{x+t} = \log\left(x+t\right)|_0^1 = \log\left(1+\frac{1}{x}\right)

Also, we can compute the above sum by noticing that 1A=0eAx dx\frac{1}{A} = \int_0^\infty e^{-Ax}\ dx

Thus limnr=1n1nx+r=limnr=1n0e(nx+r)t dt\begin{aligned} \lim_{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{nx+r} & = & \lim_{n \rightarrow \infty} \sum_{r=1}^n \int_0^\infty e^{-(nx+r)t}\ dt \\ \end{aligned} Because the integral is independent of the sum, we can interchange the sum and integral as limnr=1n0e(nx+r)t dt=limn0r=1n(et)renxt dt\begin{aligned} \lim_{n \rightarrow \infty} \sum_{r=1}^n \int_0^\infty e^{-(nx+r)t}\ dt & = & \lim_{n \rightarrow \infty} \int_0^\infty \sum_{r=1}^n (e^{-t})^{r} e^{-nxt}\ dt \\ \end{aligned} Setting y=ety = e^{-t} yields, limn0r=1n(et)renxt dt=limn01r=1nyr ynx dyy=limn01yn1y1ynx dy=limn01ynx+nynxy1 dy\begin{aligned} \lim_{n \rightarrow \infty} \int_0^\infty \sum_{r=1}^n (e^{-t})^{r} e^{-nxt}\ dt & = & \lim_{n \rightarrow \infty} \int_0^1 \sum_{r=1}^n y^r\ y^{nx}\ \frac{dy}{y} \\ & = & \lim_{n \rightarrow \infty} \int_0^1 \frac{y^n-1}{y-1} y^{nx}\ dy \\ & = & \lim_{n \rightarrow \infty} \int_0^1 \frac{y^{nx+n}-y^{nx}}{y-1}\ dy \\ \end{aligned} Separate the above sum as limn01ynx+nynxy1 dy=limn(01ynx+n1y1 dy01ynx1y1 dy)\begin{aligned} \lim_{n \rightarrow \infty} \int_0^1 \frac{y^{nx+n}-y^{nx}}{y-1}\ dy = \lim_{n \rightarrow \infty} \left( \int_0^1 \frac{y^{nx+n}-1}{y-1}\ dy - \int_0^1 \frac{y^{nx}-1}{y-1}\ dy \right) \\ \end{aligned} Now, remember that by the definition of digamma function, we have ψ(s+1)=γ+01ts1t1 dt\psi(s+1) = -\gamma + \int_0^1 \frac{t^s-1}{t-1}\ dt

Thus, we can conclude that limnr=1n1nx+r=limnψ(nx+n+1)ψ(nx+1)=log(1+1x)\lim_{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{nx+r} = \lim_{n \rightarrow \infty} \psi(nx+n+1) - \psi(nx+1) = \log\left( 1+\frac{1}{x}\right)

Proof 2

My other proof was exactly same as Ronak's, which makes use of recurrence function of digamma function.

Furthermore, I make a conjecture here that the above property holds true for some complex xx too but I'm still working on it's proof. Any help would be appreciated.

Thanks,

Kishlaya Jaiswal.

Note by Kishlaya Jaiswal
4 years, 8 months ago

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Proof is quite simple (simple properties of digamma functions)

We being with the property :

ψ(y+1)=1y+ψ(y)\displaystyle \psi{(y+1)}=\dfrac{1}{y} + \psi{(y)}

Put y=nx+1y=nx+1 to get :

ψ(nx+2)=1nx+1+ψ(nx+1) \displaystyle \psi{(nx+2)}= \dfrac{1}{nx+1} + \psi{(nx+1)}

ψ(nx+3)=1nx+2+ψ(nx+2)\Rightarrow \displaystyle \psi{(nx+3)}= \dfrac{1}{nx+2} + \psi{(nx+2)}

Continuing like this we finally have :

ψ(nx+n+1)=r=1n1nx+r+ψ(nx+1) \displaystyle \psi{(nx+n+1)}= \sum _{ r=1 }^{ n }{ \frac { 1 }{ nx+r } } + \psi{(nx+1)}

Finally we have :

r=1n1nx+r=ψ(nx+n+1)ψ(nx+1)\displaystyle \sum _{ r=1 }^{ n }{ \frac { 1 }{ nx+r } } = \psi{(nx+n+1)} - \psi{(nx+1)}

Now I am using another property converting riemann sum into an integral :

limn1nr=1nf(rn)=01f(x)dx \displaystyle \lim _{ n\rightarrow \infty }{ \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ f\left(\frac { r }{ n } \right) } } = \int _{ 0 }^{ 1 }{ f(x)dx }

Using this property we have :

limn1nr=1n1x+rn=01dtx+t \displaystyle \lim _{ n\rightarrow \infty }{\frac { 1 }{ n } \sum _{ r=1 }^{ n }{ \frac { 1 }{ x+\frac { r }{ n } } }} =\int _{ 0 }^{ 1 }{ \frac { dt }{ x+t } }

Finally getting :

limnψ(nx+n+1)ψ(nx+1)=log(1+1x) \displaystyle \lim _{ n\rightarrow \infty }{ \psi (nx+n+1)-\psi (nx+1) } =log\left(1+\dfrac { 1 }{ x } \right)

Ronak Agarwal - 4 years, 8 months ago

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I got to that one till the sum but was very confused on how to get that sum. Nice, that was new to me! I didn't know about that conversion until now. Thanks!

Kartik Sharma - 4 years, 8 months ago

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Shame on me , I was working on it from a totally different perspective .

Short and Sweet solution @Ronak Agarwal

A Former Brilliant Member - 4 years, 8 months ago

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Here's what I did :

As we know, ψ(p+1)=Hpγ\psi(p+1)=H_{p}-\gamma.

Also, Hp=r=1p1r=01xp1x1dxH_{p}=\displaystyle\sum_{r=1}^{p} \dfrac{1}{r}=\displaystyle\int_{0}^{1}\dfrac{x^p-1}{x-1}\mathrm{d}x.

So making use of the above equalities, our expression becomes ψ(nx+n+1)ψ(nx+1)=limn01tnx+n1t1dt01tnx1t1dt=limn01tnx1t(tn1)t1dt=limn01tnx1r=1ntrdt=limnr=1n01tnx+r1dt=limnr=1n1nx+r=limn1nr=1n1x+r/n=011x+ydy=log(1+x)log(x)=log(1+1x)\begin{aligned} \psi(nx+n+1)-\psi(nx+1)&=\lim_{n \to \infty}\int_{0}^{1} \dfrac{t^{nx+n}-1}{t-1}\mathrm{d}t-\int_{0}^{1} \dfrac{t^{nx}-1}{t-1}\mathrm{d}t\\ &=\lim_{n \to \infty} \int_{0}^{1}t^{nx-1}\dfrac{t(t^n-1)}{t-1}\mathrm{d}t\\ &=\lim_{n \to \infty}\int_{0}^{1}t^{nx-1}\sum_{r=1}^{n}t^r \mathrm{d}t\\ &= \lim_{n \to \infty}\sum_{r=1}^{n}\int_{0}^{1}t^{nx+r-1}\mathrm{d}t\\ &=\lim_{n \to \infty}\sum_{r=1}^{n}\dfrac{1}{nx+r}\\ &=\lim_{n \to \infty}\dfrac{1}{n}\sum_{r=1}^{n}\dfrac{1}{x+r/n}\\ &=\int_{0}^{1} \dfrac{1}{x+y} \mathrm{d}y\\ &=\log(1+x)-\log(x)\\ &=\boxed{\log\left(1+\dfrac{1}{x}\right)}\end{aligned}

NOTE : HpH_p is the pthp^{\text{th}} harmonic number.

Pratik Shastri - 4 years, 8 months ago

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Hi Pratik , what's with you changing your profile pic . Don't like Federer anymore ?

A Former Brilliant Member - 4 years, 8 months ago

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I'd never stop liking Federer :) I changed it just like that :P

Pratik Shastri - 4 years, 8 months ago

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@Pratik Shastri :)

A Former Brilliant Member - 4 years, 8 months ago

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Nice observation skills , Kishlaya Jaiswal :)

We know that ψ(z+1)=F(x) \psi (z+1) = F(x)

So using it , limnψ(nx+n+1)ψ(nx+1)=F(nx+n)F(nx)=ddx(ln((nx+n)!))ddx(ln((nx)!))=ddxln((nx+n)!(nx)!)\lim_{n \rightarrow \infty} \psi(nx+n+1)-\psi(nx+1) \\= F(nx + n) - F(nx) \\= \frac{d}{dx} ( ln((nx+n)!)) - \frac{d}{dx} (ln((nx)!)) = \frac{d}{dx} ln (\dfrac{(nx+n)!}{(nx)!})

Where ψ(x) \psi(x) is the Digamma Function and F(x)F(x) is the logarithmic derivative of the Factorial function defined as ddxln(x!)\dfrac{d}{dx} ln(x!)

Any help on how to proceed next ?

A Former Brilliant Member - 4 years, 8 months ago

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ok, here's another hint for you : try computing the sum of following series in two different ways.

limnr=0n1nx+r\lim_{n \rightarrow \infty} \sum_{r=0}^n \frac{1}{nx+r}

Kishlaya Jaiswal - 4 years, 8 months ago

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Ok, I'll try using your hint . Thanks

A Former Brilliant Member - 4 years, 8 months ago

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I too got till this one except I think it should start from r=1r=1. But I am still unable to solve the sum.

Kartik Sharma - 4 years, 8 months ago

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Is someone trying this problem, or should I post the proof?

Kishlaya Jaiswal - 4 years, 8 months ago

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Sorry , I couldn't solve it past my initial efforts despite your Hint . BTW Wait for some hours, I'll reshare this note again ,I guess not many people have seen this note .

Looking forward to your proof ¨\ddot\smile

A Former Brilliant Member - 4 years, 8 months ago

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ok, as you say. And thanks for resharing it.

Therefore, I guess, I'll post the proof within next 12 hours. :):). Will that be fine?

Kishlaya Jaiswal - 4 years, 8 months ago

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@Kishlaya Jaiswal Your Wish . Actually the next time I'm on Brilliant will be at night, so I guess it's fine by me ¨\ddot\smile

A Former Brilliant Member - 4 years, 8 months ago

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I am trying this.

Ronak Agarwal - 4 years, 8 months ago

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It must be applicable to complex numbers as well since the property of digamma function I used in the proof holds for complex numbers as well( you can always add 1 to complex number as well and hence create the summation)

Also when I converted riemann sum(right) into an integral the evaluation of that integral is governed by fundamental theorom of calculus and since the anti-derivative I calculated holds for complex numbers as well hence I believe your result is justified for complex numbers as well.

But since I am not being rigourous hence I may not be sure( If I am missing something)

Ronak Agarwal - 4 years, 8 months ago

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Ok, done! I've added my proof also.

Kishlaya Jaiswal - 4 years, 8 months ago

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As usual , your methods are the best :) Thanks and btw I realized the error in my calculations but yours and Pratik's methods are the best . I had tried Pratik's method just before I used the one that I posted . That one seemed to be so close to the final result that I proceeded with it . Can you please check if a proof using it is possible ? Personally, with my limited knowledge I don't think it's possible , but I just can't help stating that it looks quite similar to the final answer .

Thanks for the same :)

A Former Brilliant Member - 4 years, 8 months ago

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