Bernoulli Numbers - Explanation

Augusta Ada King-Noel, Countess of Lovelace (10 December 1815 – 27 November 1852) was an English mathematician and writer, chiefly known for her work on Charles Babbage’s proposed mechanical general-purpose computer, the Analytical Engine. [1].
She is widely regarded as the first computer programmer. She wrote an algorithm to calculate the Bernoulli numbers, for more please visit the previous blog here.
So, I thought to study and analyze Bernoulli numbers. $\mathcal Crazy \ math \ ahead!$
$B_0 = 1; B_1= \pm \frac{1}{2}; B_2 = \frac{1}{6}; B_3 = 0; B_4=-\frac{1}{30}; B_5 = 0...$

Approach 1 :

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Let us begin,
Definition: The Bernoulli numbers are defined as the co-efficients of the power series of the expansion of $\large \frac{t}{e^{t} -1}$. For $m \geq 0$, we define $B_m$ so that,
$$\large \frac{t}{e^{t} -1} = \sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m}-(1)$$


Let us apply some mathematical rigor into it and see what happens.
We have,
$$\large \frac{t}{e^{t} -1} = \sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m}\\ \implies\large (e^{t} -1) \frac{t}{e^{t} -1} = (e^{t} -1)\sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m} \\ \implies\large {t}= (e^{t} -1)\sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m} -(2)$$
Now, we know the McLaurin Series is,
$f(t) = f(0) + f'(0) + \frac{f""(0)}{2!}t^{2} + \frac{f^{(3)}(0)}{3!}t^{3}+...+\frac{f^{(n)} }{n!}t^{n}$
Let, $f(t) = e^{t}$, to find its McLaurin coefficients we must evaluate $\large(\frac{d^{k}}{dt^{k}}f(t))|_{t=0}$ for all $k$ = $0, 1, 2, ...$

By substitution, the McLaurin Series expansion of $(e^{t})$ is,
$$\large \sum_{k=0}^{\infty}\frac{f^{k}(0)}{k!}(t-0)^{k} = \frac{x^{0}}{0!} + \frac{x^{1}}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ...\\ \implies \large e^{t} = \frac{x^{0}}{0!} + \frac{x^{1}}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ...\\ \implies \large e^{t} = 1 + \frac{x^{1}}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ...\\ \implies \large (e^{t} -1) = \frac{x^{1}}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ...\\ \implies \large (e^{t} -1) = \sum_{k=1}^{\infty}\frac{t^{k}}{k!} - (3)$$

Using $(3)$ in $(2)$, we get,
$$\implies\large {t}= \sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m} .\sum_{k=1}^{\infty}\frac{t^{k}}{k!} \\ \implies\large {t}= \sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m} .\sum_{k=1}^{\infty}\frac{t^{k}}{k!} \\ \implies\large \frac{t}{t}= \frac{1}{t}\sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m} .\sum_{k=1}^{\infty}\frac{t^{k}}{k!} \\ \implies\large \frac{t}{t}= \sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m} .\sum_{k=1}^{\infty}\frac{t^{k-1}}{k!} \\ \implies\large 1= \sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m} .\sum_{k=1}^{\infty}\frac{t^{k}}{(k+1)!} - (4)$$
Let’s see which coefficient $a_n$ has the $t^n$ in the expanded right-hand part of this equation for some $n > 0$ (we expect this coefficient to be a zero). $t^n$ may appear if $t^m$ from the first sum multiplies the $t^{n-m}$ from the second one for some $0\le m\le n$. Thus,
$$\large a_n = \sum_{m=0}^{n}\frac{B_{n-k}}{m!(n-m+1)!}\\ \large = \sum_{m=0}^{n} {n+1\choose m} \frac{B_{m}}{(n+1)!}\\ \large = \sum_{m=0}^{n-1} {n+1\choose m} \frac{B_{m}}{(n+1)!} + \\\binom{n+1}{n}\frac{B_n}{(n+1)!} = 0$$
for any $n > 0$. Thus
$$\large (n+1)B_n = -\sum_{m=0}^{n-1} {n+1\choose m} \frac{B_{m}}{(n+1)!}$$
Now for $n = 0$ we have from $(1)$
$$\large \frac{B_0}{0!}\frac{1}{1!} = 1$$
and thus
$$B_0 = 1$$

Approach 2 :

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We know,
$$\large \sum_{i=1}^{n} (i) = 1 + 2 + 3 + ... = \frac{n(n+1)}{2},\\ \large \sum_{i=1}^{n} (i^{2}) = 1^{2} + 2^{2} + 3^{2} + ... = \frac{n(n+1)(2n+1)}{6},\\ \sum_{i=1}^{n} (i^{3}) = 1^{3} + 2^{3} + 3^{3} + ...=\frac{n^{2}(n+1)^{2}}{4},\\ \sum_{i=1}^{n} (i^{4}) = 1^{4} + 2^{4} + 3^{4} + ...=\frac{n(n+1)(2n+1)(3n^{2}+3n-1}{30}.$$
and so on…

Then, how about,
$$\large \sum_{i=1}^{\infty}(i^{p})=?$$
Mathematicians have always been fascinated with such classic general formulae. So was JohannFaulhaber.
Let, $$S_p(n) = \sum_{i=1}^{\infty}(i^{p}) \\ \forall p \geq 0$$
Define the following exponential generating function with (initially) indeterminate $z$
$G(z,n) = \sum_{p=0}^{\infty } S_{p}(n) \frac {1}{p!}z^{p}$
We find
$$G(z,n)=\sum _{p=0}^{\infty }\sum _{k=1}^{n}\frac {1}{p!}((kz)^{p} )\\ =\sum _{k=1}^{n}e^{kz} \\ =e^{z}.\frac {1-e^{nz}}{1-e^{z}} \\ ={\frac {1-e^{nz}}{e^{-z}-1}}. \\ $$

$$G(z,n) =\sum_{p=0}^{\infty} \sum_{k=1}^{n} \frac{1}{p!}(kz)^p\\ =\sum_{k=1}^{n}e^{kz}=e^{z}.\frac{1-e^{nz}}{1-e^{z}},\\ = \frac{1-e^{nz}}{e^{-z}-1}.$$
This is an entire function in $z$ so that $z$ can be taken to be any complex number.
We next recall the exponential generating function for the Bernoulli polynomials $B_{j}(x) B_j(x)$
$${\frac {ze^{zx}}{e^{z}-1}}=\sum _{j=0}^{\infty }B_{j}(x)\frac{z^{j}}{j!}$$
where $B_{j}=B_{j}(0)$ denotes the Bernoulli number (with the convention $B_{1}=\frac {1}{2}; B_{1}=-\frac{1}{2}$). We obtain the Faulhaber formula by expanding the generating function as follows:
$$G(z,n) = \sum_{j=0}^{\infty}B_j \frac{(-z)^{j-1}}{j!}(-\sum_{l=1}^{\infty}\frac{(nz)^l}{l!}\\ $$
Solving it is again very complex, so, finally we get,
$$\large \sum_{k=1}^{n}i^{p}=\frac{1}{p+1}\sum_{j=0}^{p}(-1)^{j}\binom{p+1}{j}B_jn^{p+1-j}$$
Note that, $B_j = 0$, $\forall$ odd $j>1$; that is why Faulhaber defines $B_{1} = \frac{1}{2}$.

Verification:

Let us consider the following values,
$$p = 2 \\ $$
$$L.H.S = \large \sum_{k=1}^{n}i^{p} \\ = 1^{2} + 2^{2} + 3^{2} +...+n^{2}= \frac{n(n+1)(2n+1)}{6}.$$
$$R.H.S = \frac{1}{2+1}\sum_{j=0}^{2}(-1)^{j}\binom{2+1}{j}B_jn^{2+1-j}\\ =\frac{1}{2+1}[(-1)^{0}\binom{3}{0}B_0n^{3-0}+(-1)^{1}\binom{3}{1}B_1n^{3-1}+\\(-1)^{2}\binom{3}{2}B_2n^{3-2}]\\ =\frac{1}{3}[1 \cdot 1\cdot B_0 \cdot n^{3}+(-1) \cdot 3\cdot B_1 \cdot n^{2}+\\1 \cdot 3 \cdot B_2 \cdot n^{1}]\\ $$
We put, $B_0=1$ , $B_1=\frac{1}{2}$ and $B_2=\frac{1}{6}$ and see that $LHS$ = $RHS$.

Hence, Proved.