Go Uniformly

First let us evaluate the limit $\large \lim_{n\to\infty} g(n,k) = \lim_{n\to\infty} (\int_{0}^{1}(1-x^{k})^{n}dx)^{\frac{1}{n}}$

Substitute $x=t^{\frac{1}{k}}$ .

So we get $\large \lim_{n\to\infty} (\frac{1}{k}\int_{0}^{1}t^{\frac{1}{k}-1}(1-t)^{n}dt)^{\frac{1}{n}}$

Using Beta Function we get

$\large \lim_{n\to\infty} (\frac{1}{k}\Beta(\frac{1}{k},n+1))^{\frac{1}{n}}$

$\large = \lim_{n\to\infty} (\frac{1}{k}\frac{\Gamma(\frac{1}{k})\Gamma(n+1)}{\Gamma(n+1+\frac{1}{k})})^{\frac{1}{n}}$

Now here comes some tricky parts.

We see that $\large \Gamma(\frac{1}{k})$ takes finite values for all $k \in \mathbb{N}$ .

So $\large \Gamma(\frac{1}{k})^{\frac{1}{n}}$ tends to $1$ as $n\to\infty$ . And also $\large (\frac{1}{k})^{\frac{1}{n}}$ tends to $1$ as $n\to\infty$ .

So we can approximate $\large \lim_{n\to\infty} g(n,k)$ as :-

$\large = \lim_{n\to\infty} (\frac{\Gamma(n+1)}{\Gamma(n+1+\frac{1}{k})})^{\frac{1}{n}}$

Now we want to use Stirling's approximation for $\Gamma(x+1)$

Using that we get $\large\large = \lim_{n\to\infty} (\frac{(\frac{n}{e})^{n}\sqrt{2n\pi}}{(\frac{(n+\frac{1}{k})}{e})^{(n+\frac{1}{k})}\sqrt{2(n+\frac{1}{k})\pi}})^{\frac{1}{n}}$

Now again we want to use another approximation

We see that $(\sqrt{2n\pi})^{\frac{1}{n}}$ and $(\sqrt{2(n+\frac{1}{k})\pi})^{\frac{1}{n}}$ tend to $1$ as $n\to\infty$ . Proof of this is elementary and can be found in any real analysis book. So we can approximate $\large \lim_{n\to\infty} g(n,k)$ as :-

$\large \lim_{n\to\infty}\frac{\frac{n}{e}}{\frac{n+\frac{1}{k}}{e}} = 1$

Here is a graph of this. As $k$ increases it approaches $1$ faster and faster as $n \to \infty$

Now we have $\large \lim_{M\to\infty} \sum_{N=1}^{M}(\sum_{m=1}^{N}\sum_{k=1}^{m}\frac{1}{m^{2}} - \ln(N+1))\frac{e}{M}$

$\large = \lim_{M\to\infty} \sum_{N=1}^{M}(\sum_{m=1}^{N}\frac{1}{m} - \ln(N+1))\frac{e}{M}$

$\large = \lim_{M\to\infty} \sum_{N=1}^{M}(H_{N} - \ln(N+1))\frac{e}{M}$

where $H_{n}$ denotes the nth harmonic number .

Now I would like to state a theorem which is less used , but when it is indeed used...It is absolutely a beast. The Theorem is known as Cauchy's First Limit Theorem. Which basically states that:-

If $\{x_{n}\}_{n}$ be a sequence of real numbers and if $\large \lim_{n\to\infty} \{x_{n}\} = l$ . Then the arithmetic mean(as well as the geometric mean) of all the terms of $\{x_{n}\}$ converges to the same limit $l$ as $n\to \infty$ . Which is expressed as $\large \lim_{n\to\infty}\frac{\sum_{k=1}^{n} x_{k}}{n} = l = \lim_{n\to\infty} \sqrt[n]{\prod_{k=1}^{n} x_{k}}$ .

So let us call $\large \{x_{r}\}_{r} =(H_{r} - \ln(r+1))$ . Now we know that $\large \lim_{r\to\infty} x_{r}$ converges to $\gamma$ . Where $\gamma$ denotes the Euler Mascheroni Constant .

So by Cauchy's First Limit Theorem we have $\large \lim_{M\to\infty}\frac{\sum_{k=1}^{M} x_{k}}{M}= \gamma$ .

So we have our answer as $e \cdot \gamma \approx 1.569$ .

Proof of Cauchy's First limit theorem:- https://brilliant.org/discussions/thread/cauchys-first-limit-theorem/