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 \(x > 0\)

lim⁡n→∞ψ(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 lim⁡n→∞∑r=1n1nx+r\lim_{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{nx+r}

We can convert the above riemann sum into integral as :

lim⁡n→∞∑r=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=∫0∞e−Ax dx\frac{1}{A} = \int_0^\infty e^{-Ax}\ dx

Thus lim⁡n→∞∑r=1n1nx+r=lim⁡n→∞∑r=1n∫0∞e−(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 lim⁡n→∞∑r=1n∫0∞e−(nx+r)t dt=lim⁡n→∞∫0∞∑r=1n(e−t)re−nxt 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=e−ty = e^{-t} yields, lim⁡n→∞∫0∞∑r=1n(e−t)re−nxt dt=lim⁡n→∞∫01∑r=1nyr ynx dyy=lim⁡n→∞∫01yn−1y−1ynx dy=lim⁡n→∞∫01ynx+n−ynxy−1 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 lim⁡n→∞∫01ynx+n−ynxy−1 dy=lim⁡n→∞(∫01ynx+n−1y−1 dy−∫01ynx−1y−1 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+∫01ts−1t−1 dt\psi(s+1) = -\gamma + \int_0^1 \frac{t^s-1}{t-1}\ dt

Thus, we can conclude that lim⁡n→∞∑r=1n1nx+r=lim⁡n→∞ψ(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.

#Calculus #Convergence #Properties #Kishlaya #DigammaFunctions

Note by Kishlaya Jaiswal
5 years, 7 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 :

lim⁡n→∞1n∑r=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 :

lim⁡n→∞1n∑r=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 :

lim⁡n→∞ψ(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 - 5 years, 7 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 - 5 years, 7 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 - 5 years, 7 months ago

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Here's what I did :

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

Also, Hp=∑r=1p1r=∫01xp−1x−1dxH_{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)=lim⁡n→∞∫01tnx+n−1t−1dt−∫01tnx−1t−1dt=lim⁡n→∞∫01tnx−1t(tn−1)t−1dt=lim⁡n→∞∫01tnx−1∑r=1ntrdt=lim⁡n→∞∑r=1n∫01tnx+r−1dt=lim⁡n→∞∑r=1n1nx+r=lim⁡n→∞1n∑r=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 - 5 years, 7 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 - 5 years, 7 months ago

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

Pratik Shastri - 5 years, 7 months ago

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@Pratik Shastri – :)

A Former Brilliant Member - 5 years, 7 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 , lim⁡n→∞ψ(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 - 5 years, 7 months ago

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

lim⁡n→∞∑r=0n1nx+r\lim_{n \rightarrow \infty} \sum_{r=0}^n \frac{1}{nx+r}

Kishlaya Jaiswal - 5 years, 7 months ago

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Ok, I'll try using your hint . Thanks

A Former Brilliant Member - 5 years, 7 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 - 5 years, 7 months ago

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

Kishlaya Jaiswal - 5 years, 7 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 - 5 years, 7 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 - 5 years, 7 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 - 5 years, 7 months ago

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I am trying this.

Ronak Agarwal - 5 years, 7 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 - 5 years, 7 months ago

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

Kishlaya Jaiswal - 5 years, 7 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 - 5 years, 7 months ago

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