Convergence of Digamma Functions.

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 · 2 years, 5 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 \[\begin{align} \psi(nx+n+1)-\psi(nx+1)&=\lim_{n \to \infty}\int_{0}^{1} \dfrac{t^{nx+n}-1}{t-1}\mathrm{d}t-\int_{0}^{1} \dfrac{t^{nx}-1}{t-1}\mathrm{d}t\\ &=\lim_{n \to \infty} \int_{0}^{1}t^{nx-1}\dfrac{t(t^n-1)}{t-1}\mathrm{d}t\\ &=\lim_{n \to \infty}\int_{0}^{1}t^{nx-1}\sum_{r=1}^{n}t^r \mathrm{d}t\\ &= \lim_{n \to \infty}\sum_{r=1}^{n}\int_{0}^{1}t^{nx+r-1}\mathrm{d}t\\ &=\lim_{n \to \infty}\sum_{r=1}^{n}\dfrac{1}{nx+r}\\ &=\lim_{n \to \infty}\dfrac{1}{n}\sum_{r=1}^{n}\dfrac{1}{x+r/n}\\ &=\int_{0}^{1} \dfrac{1}{x+y} \mathrm{d}y\\ &=\log(1+x)-\log(x)\\ &=\boxed{\log\left(1+\dfrac{1}{x}\right)}\end{align}\]

NOTE : \(H_p\) is the \(p^{\text{th}}\) harmonic number. – Pratik Shastri · 2 years, 5 months ago

Ok, done! I've added my proof also. – Kishlaya Jaiswal · 2 years, 5 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) – Ronak Agarwal · 2 years, 5 months ago

Is someone trying this problem, or should I post the proof? – Kishlaya Jaiswal · 2 years, 5 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 ? – Azhaghu Roopesh M · 2 years, 5 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 :) – Azhaghu Roopesh M · 2 years, 5 months ago