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.

Markdown

Appears as

*italics* or _italics_

italics

**bold** or __bold__

bold


- bulleted
- list

bulleted
list


1. numbered
2. list

numbered
list

Note: you must add a full line of space before and after lists for them to show up correctly

paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)

example link

> This is a quote

This is a quote

    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"

# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"

Math

Appears as

Remember to wrap math in $ ... $ or \[ ... \] to ensure proper formatting.

2 \times 3

2 \times 3

2^{34}

2^{34}

a_{i-1}

a_{i-1}

\frac{2}{3}

\frac{2}{3}

\sqrt{2}

\sqrt{2}

\sum_{i=1}^3

\sum_{i=1}^3

\sin \theta

\sin \theta

\boxed{123}

\boxed{123}

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Sort by:

Top Newest

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 :

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

Ronak Agarwal - 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!

Kartik Sharma - 5 years, 3 months ago

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, 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.

Pratik Shastri - 5 years, 3 months ago

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

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

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 ?

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

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

Kishlaya Jaiswal - 5 years, 3 months ago

Ok, I'll try using your hint . Thanks

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

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

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 $\ddot\smile$

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

I am trying this.

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)

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

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 :)

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$