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I found the following formula here: Taylor Series of $\tan x$.

Taylor series of $\tan x$:

$$\tan x = \sum_{n\,=\,1}^\infty \frac {(-1)^{n-1}2^{2n} (2^{2n}-1) B_{2n}} {(2n)!} x^{2n - 1} $$.

I do not understand what the Beournoulli number means in this formula. My problems are the following ones:

I do not know anything about bernoulli numbers. I do not know how to expand the bernoulli number in the formula you see further up. What expansion will be the final result? How did you arrive there? Why did you arrive there? How did you know that you should use the specific bernoulli-related terms that you used in your answer? What is the purpose of the bernoulli number in this formula?

A great deal of thanks.

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    The Bernoulli numbers appear from Riemann zeta function. Do you about it? – 2017-01-15
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    @MyGlasses Nope. – 2017-01-15
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    @MyGlasses But you could make a explanation with it - I can make some research on the Riemann zeta func. – 2017-01-15
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    The best book, that I know, about _Bernoulli numbers_, _Riemann zeta function_ is **Apostol Tom Mike, Introduction to Analytic Number Theory, Springer-Verlag, New York, (1976)(352s)** , pp266-270. – 2017-01-15
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    @MyGlasses https://i.stack.imgur.com/ATuev.png Is this the book that you mean? Is it the first or the second volume that you are referring to? – 2017-01-15
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    Yes. Volume one I mean. Volume two speaks about pure number theory. It has another name i don't remember – 2017-01-15
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    @MyGlasses That book is to far beyond my current level of mathematical understanding. Am I screwed in understanding the example of the taylor series of tan x, that you see further up, if so? It should not be the case. – 2017-01-15
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    Where will you use $\tan $ series, if possible to say. – 2017-01-15
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    @MyGlasses I don't know really, I just found the series particularly interesting. – 2017-01-15
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    Oh, sorry. I see that you did present some context. – 2017-01-17

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The exponential generating function for the Bernoulli Numbers is $$ \sum_{n=0}^\infty\frac{B_nx^n}{n!}=\frac{x}{e^x-1}\tag{1} $$ The even part of $(1)$ is $$ \sum_{n=0}^\infty\frac{B_{2n}x^{2n}}{(2n)!}=\frac x2\coth\left(\frac x2\right)\tag{2} $$ Since $\cot(x)=i\coth(ix)$, by substituting $x\mapsto ix$, we get $$ \begin{align} \sum_{n=0}^\infty\frac{(-1)^nB_{2n}x^{2n}}{(2n)!} &=\frac{ix}2\coth\left(\frac{ix}2\right)\\ &=\frac x2\cot\left(\frac x2\right)\tag{3} \end{align} $$ Therefore, $$ \cot(x)=\sum_{n=0}^\infty\frac{(-1)^nB_{2n}2^{2n}x^{2n-1}}{(2n)!}\tag{4} $$ Since $\tan(x)=\cot(x)-2\cot(2x)$, we get $$ \begin{align} \tan(x) &=\sum_{n=0}^\infty\frac{(-1)^nB_{2n}2^{2n}\left(x^{2n-1}-2\cdot2^{2n-1}x^{2n-1}\right)}{(2n)!}\\ &=\sum_{n=1}^\infty\frac{(-1)^{n-1}B_{2n}\left(2^{2n}-1\right)2^{2n}x^{2n-1}}{(2n)!}\tag{5} \end{align} $$