# 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 $$x > 0$$

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

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 $\lim_{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{nx+r}$

We can convert the above riemann sum into integral as :

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

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

Thus \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 \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 $y = e^{-t}$ yields, \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 \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 $\psi(s+1) = -\gamma + \int_0^1 \frac{t^s-1}{t-1}\ dt$

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

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 $x$ too but I'm still working on it's proof. Any help would be appreciated.

Thanks,

Kishlaya Jaiswal.

Note by Kishlaya Jaiswal
5 years, 3 months ago

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Proof is quite simple (simple properties of digamma functions)

We being with the property :

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

Put $y=nx+1$ to get :

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

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

Continuing like this we finally have :

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

Finally we have :

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

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

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

Using this property we have :

$\displaystyle \lim _{ n\rightarrow \infty }{\frac { 1 }{ n } \sum _{ r=1 }^{ n }{ \frac { 1 }{ x+\frac { r }{ n } } }} =\int _{ 0 }^{ 1 }{ \frac { dt }{ x+t } }$

Finally getting :

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

- 5 years, 3 months ago

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!

- 5 years, 3 months ago

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

Short and Sweet solution @Ronak Agarwal

- 5 years, 3 months ago

Here's what I did :

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

Also, $H_{p}=\displaystyle\sum_{r=1}^{p} \dfrac{1}{r}=\displaystyle\int_{0}^{1}\dfrac{x^p-1}{x-1}\mathrm{d}x$.

So making use of the above equalities, our expression becomes \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 : $H_p$ is the $p^{\text{th}}$ harmonic number.

- 5 years, 3 months ago

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

- 5 years, 3 months ago

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

- 5 years, 3 months ago

:)

- 5 years, 3 months ago

Nice observation skills , Kishlaya Jaiswal :)

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

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

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

Any help on how to proceed next ?

- 5 years, 3 months ago

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

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

- 5 years, 3 months ago

Ok, I'll try using your hint . Thanks

- 5 years, 3 months ago

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

- 5 years, 3 months ago

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

- 5 years, 3 months ago

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 $\ddot\smile$

- 5 years, 3 months ago

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?

- 5 years, 3 months ago

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

- 5 years, 3 months ago

I am trying this.

- 5 years, 3 months ago

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)

- 5 years, 3 months ago

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

- 5 years, 3 months ago

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 :)

- 5 years, 3 months ago