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I was trying to understand Bernoulli numbers. When I googled, I found this link.

It starts by saying that, The Bernoulli numbers are defined via the coefficients of the power series expansion of $\frac{t}{e^{t}-1}$,

then they write the expansion,for $m \geq 0$.

$\frac{t}{e^{t}-1} = \sum_{m=0}^{\infty} \frac{B_m}{m!}t^{m}$ (1)

Then, the tutorial says that we multiply both sides by $(e^{t} -1)$ and get

$B_0=1$,

$B_m= -\frac{1}{m+1}\sum_{k=0}^{m-1}\binom{m+1}{k}B_k$. (2)

How is this happening?

I tried doing this with hand and I get a recursive relation, that looks like,

$\frac{t}{e^{t}-1} = \sum_{m=0}^{\infty} \frac{B_m}{m!}t^{m}$

$\implies t = e^{t}\sum_{m=0}^{\infty} \frac{B_m}{m!}t^{m} - \sum_{m=0}^{\infty} \frac{B_m}{m!}t^{m}$

$\implies t = e^{t}\sum_{m=0}^{\infty} \frac{B_m}{m!}t^{m} - \frac{t}{e^{t}-1}$

What next?

How do I get to (2)?

I am new to number theory, please excuse me if the doubt is very basic. Any kind of help will be appreciated. :)

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2 Answers 2

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Using Maclaurin series for $e^t$ we get that $$ e^t - 1 = \sum_{k=1}^{\infty}\frac{t^k}{k!}. $$ Now we get $$ t = \left(\sum_{m=0}^{\infty}\frac{B_m}{m!}t^m\right)\left(\sum_{k=1}^{\infty}\frac{1}{k!}t^k\right) $$ or, dividing by $t$ $$ 1 = \left(\sum_{m=0}^{\infty}\frac{B_m}{m!}t^m\right)\left(\sum_{k=1}^{\infty}\frac{1}{k!}t^{k-1}\right) = \left(\sum_{m=0}^{\infty}\frac{B_m}{m!}t^m\right)\left(\sum_{k=0}^{\infty}\frac{1}{(k+1)!}t^{k}\right). \tag1 $$ 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 \begin{align} a_n = \sum_{m=0}^{n}\frac{B_{n-k}}{m!(n-m+1)!}= \sum_{m=0}^{n} {n+1\choose m} \frac{B_{m}}{(n+1)!} = \\ = \sum_{m=0}^{n-1} {n+1\choose m} \frac{B_{m}}{(n+1)!} + {n+1 \choose n}\frac{B_n}{(n+1)!} = 0 \end{align} for any $n > 0$. Thus $$ (n+1)B_n = -\sum_{m=0}^{n-1} {n+1\choose m} \frac{B_{m}}{(n+1)!} $$ which implies result $(2)$ from your tutorial (reassign $n\to m$, $m\to k$).

Now for $n = 0$ we have from $(1)$ $$ \frac{B_0}{0!}\frac{1}{1!} = 1 $$ and thus $B_0 = 1$.

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  • $\begingroup$ WOW! Thank you for this. Really very clear with it now. $\endgroup$ Mar 29, 2017 at 13:30
  • $\begingroup$ What books did you refer to be this good at Math? Can you give me some names? $\endgroup$ Mar 29, 2017 at 13:31
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With recurrences of this type I like the representation in a matrix format. Note, that the binomial-coefficients in your recursive formula occur in the manner of the lower triangular Pascal-matrix ( = "P") and that we need a modified/trimmed version (= "Q" ) of it.

The following is the multiplication-scheme for the Bernoulli-numbers [update] corrected :

        |                               |      1  |
        |                               |   -1/2  |  vector B of
        |                               |    1/6  |  Bernoulli-numbers 
        |  -dZ * Q * B = B'             |      0  |
        |                               |  -1/30  |
        |                               |      0  |
        |                               |   1/42  |
        |                               |      0  |
     -  +  -  -   -   -   -   -   -  -  +      -  +
    -Z  |  modified Pascalmatrix Q      |         |
     -  +  -  -   -   -   -   -   -  -  +      -  +
    -1  |  .  .   .   .   .   .   .  .  |      0  |  reduced (first entry =0)
  -1/2  |  1  .   .   .   .   .   .  .  |   -1/2  |  vector B' of 
  -1/3  |  1  3   .   .   .   .   .  .  |    1/6  |  Bernoulli-numbers 
  -1/4  |  1  4   6   .   .   .   .  .  |      0  |  
  -1/5  |  1  5  10  10   .   .   .  .  |  -1/30  | 
  -1/6  |  1  6  15  20  15   .   .  .  |      0  | 
  -1/7  |  1  7  21  35  35  21   .  .  |   1/42  |
  -1/8  |  1  8  28  56  70  56  28  .  |      0  |
     -  +  -  -   -   -   -   -   -  -  +      -  +

This scheme illustrates the matrix-product $$ -\,^dZ \cdot Q \cdot B = B' $$ where $\,^d Z = \operatorname{diagonal}([1,1/2,1/3,...]) $ .
This is an explication of your formula (2) where $m$ is the matrix-row-index beginning at $0$ .
Of course: so far it is no proof, just an illustration how to understand the ingredients of formula (2). However, manipulations at this formula can be made to related it to known, simpler and proven relations between binomial-coefficients and Bernoulli-numbers, if that is what you want.

(P.s.: you might also be interested in this small treatize on exactly that subject / method)

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  • $\begingroup$ Can you start from the point where I left at (2)? Your answer explains Bernoulli Number, but not how the particular writer got the result, as stated in my question. $\endgroup$ Mar 29, 2017 at 10:09
  • $\begingroup$ Also, I would like to know how did you manage to write this answer in this particular format? Did you use some software? $\endgroup$ Mar 29, 2017 at 10:09
  • $\begingroup$ @PragyadityaDas : the sum-part in your formula (2) has the dot-product of the row of the Pascalmatrix (the sequence of $ \binom m0, \binom m1 ,...$) with the Bernoulli-numbers $b_k$ and because in (2) this is meant for any $m$ one can stack all that sum-terms below each other as the m'th row of a matrix-multiplication. But <upps> I see I have to correct my formula to relate it correctly to your (2). I missed the $ {1 \over m+1}$ - term. $\endgroup$ Mar 29, 2017 at 10:26
  • $\begingroup$ As to software: yes I use Pari/GP with a set of user-defined matrix-procedures, and a selfmade/homebrewn GUI which supports easy creation of outputs like that. (works in windows, but so far only with Pari/GP 2.2.11) If you like you can download from my homepage/software $\endgroup$ Mar 29, 2017 at 10:27

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