Do the odd coefficients of the power series for tangent form a decreasing sequence?

Question

The beginning of the power series of the tangent function is given by $$\tan(x) = x+\frac{x^3}{3}+\frac{2\,x^5}{15}+\frac{17\,x^7}{315}+ \frac{62\,x^9}{2835}+\frac{1382\,x^{11}}{155925}+\frac{21844\, x^{13}}{6081075}+\frac{929569\,x^{15}}{638512875}+\frac{6404582 \,x^{17}}{10854718875}+\frac{443861162\,x^{19}}{1856156927625} +\cdots $$

Do the odd coefficients of the series form a decreasing sequence?

It seems so. The (2n-1)th coefficient of the series is given by $$ a_n = 4^n(4^n-1)|B_{2n}|/(2n)! $$ where $B_n$ is the Bernoulli numbers. I don't know how to evaluate the quotients of the successive even Bernoulli numbers though.

Answer 1

The asymptotics of Bernoulli numbers of even index is $$B_{2 n} \sim(-1)^{n+1} \,4 \sqrt{\pi n}\, \left(\frac{n}{\pi e}\right)^{2 n}$$ (have a look here).

This is a very good a)pproximation : relative error smaller than $1.0$% for $n=5$, smaller than $0.1$% for $n=42$.

So, if

$$a_n=4^n(4^n-1)\frac{|B_{2n}|} {(2n)!}\quad \implies \quad a_n \sim \frac{4^{n-3} \left(1152 n^2-48 n+1\right) }{9 n^2 \, \pi ^{2 n}}$$ $$\frac{a_{n+1}}{a_n}=\frac{4 n^2 \left(1152 n^2+2256 n+1105\right)}{\pi ^2 (n+1)^2 \left(1152 n^2-48 n+1\right)}$$

Expanding as a Taylor series $$\frac{a_{n+1}}{a_n}=\frac 4{\pi^2} \left(1+\frac{1}{24 n^2}-\frac{1}{24n^3}+O\left(\frac{1}{n^4}\right)\right)$$

Using your last coefficients $$\frac {a_{19}}{a_{17}}=\frac{221930581}{547591761}=0.405285$$ wile the truncated series gives $$\frac {a_{19}}{a_{17}}\sim \frac{58964}{14739 \pi ^2}=0.405340$$

Edit

Interesting is the approximation

$$|B_{2 n}| \sim 4\,\sqrt{\pi n}\, \left(\frac n{\pi e}\, \frac{480 n^2+9}{480 n^2-1}\right)^{2n}$$ which is much more accurate and gives

$$\frac{a_{n+1}}{a_n}=\frac 4{\pi^2} \left(1+\frac{2369}{23224320n^6}+O\left(\frac{1}{n^7}\right)\right)$$ which gives

$$\frac {a_{19}}{a_{17}}\sim \frac{560578626480449}{140144656619520 \pi ^2}$$ which differs from the exact value by $9.32\times 10^{-10}$.

Update

If we consider instead $${a_{n+1} \over a_n} = \Bigl(4+{3 \over 4^{n}-1}\Bigr) {1 \over \pi^2} {\zeta(2n+2) \over \zeta(2n)} $$

$$ {\zeta(2n+2) \over \zeta(2n)}~>~ 4 \,\frac{2^{2 n-1}-1}{2^{2 n+1}-1}$$ would be very useful

Answer 2

I finally solved the question using a formula linked in the answer of Claude Leibovici. This formula expresses the even Bernoulli numbers in terms of the Zeta function.

$$|B_{2n}| = {(2n)!\ \zeta(2n)\over 2^{2n-1}\pi^{2n}}.$$ This gives $$a_n = {4^n-1\over \pi^{2n}} \ 2 \ \zeta(2n)$$ which in turns leads to the ratio $${a_{n+1} \over a_n} = \Bigl(4+{3 \over 4^{n}-1}\Bigr) {1 \over \pi^2} {\zeta(2n+2) \over \zeta(2n)}.$$ The number $\zeta(2n+2)$ is less than $\zeta(2n)$ because ${1\over k^{2n+2}}$ is less than ${1\over k^{2n}}$. We end with $${a_{n+1} \over a_n} \leq {5\over \pi^2} \leq 1.$$

Answer 3

We have the partial fraction expansion of $\tan x$

$$ \tan x = \sum_{n\in \mathbb{Z}} - \frac{1}{x - (n+ \frac{1}{2}) \pi}$$

Plea: Look at the Laurent expansion of $\tan x$ around a pole $(n+\frac{1}{2}) \pi$, it starts with $ - \frac{1}{x - (n+ \frac{1}{2})\pi}$. Now, we need to regularize the RHS, that is, group some terms together to get a series convergent on compacts. Grouping together symmetric terms will do. So we get the formula

$$\tan x = \sum_{n\ge 0} \left(- \frac{1}{x - (n+ \frac{1}{2}) \pi} - \frac{1}{x - (-n+ \frac{1}{2}) \pi}\right) = \sum_{n=0}^{\infty} \frac{2 x}{((n+ \frac{1}{2})\pi)^2 - x^2}$$

From this we can get the Taylor expansion of $\tan x $ at $0$. Indeed, we have $$\frac{2 x}{a^2 - x^2} = \sum_{N=0}^{\infty} \frac{2}{a^{2N+2}} x^{2N+1}$$

so

$$\tan x = \sum_{N=0}^{\infty} \left( 2 \sum_{n=0}^{\infty} \frac{1}{((n+\frac{1}{2})\pi)^{2N+2}}\right)x^{2N+1}$$

Now observe that the coefficients of $x^{2N+1}$ decrease with $N$, since all $(n+\frac{1}{2}) \pi > 1$ for $n\ge 0$.

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$\bf{Added:}$ A similar formula

$$\frac{1}{x} - \cot x = \sum_{n=1}^{\infty} \frac{2x}{(n \pi)^2 - x^2}= \sum_{N=1}^{\infty} \left ( \sum_{n=1}^{\infty} \frac{2}{(n \pi)^{2N+2}} \right) x^{2N+1}$$

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$\bf{Added}$ From the expression of the coefficients as moments, we get that $\frac{a_{n+1}}{a_n}$ is increasing, starting with $\frac{1}{3}$, and with limit $\frac{4}{\pi^2}$ ( a logarithmically convex sequence). Perhaps we can also get from this the asymptotics of $\frac{a_{n+1}}{a_n}$. (One thinks of the Euler-Maclaurin formula).

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$\bf{Added:}$ Let's record the (standard) continued fraction of $\tan x$ being

$$[0; \frac{1}{x}, -\frac{3}{x}, \frac{5}{x}, - \frac{7}{x}, \ldots]$$

The finite fraction up to $(-1)^n \frac{2n+1}{x}$ gives the Pade approximant of $\tan x$ at $0$ of order $(n+1, n)$, or $(n,n+1)$, depending on $n$ being even or odd, and so will approximate the Taylor series of $\tan x$ up to the term $\cdots x^{2n+1}$ included.

Answer 4

There is a direct solution (without previous knowledge of Bernoulli numbers and $\zeta$ function) derived from this answer obtained by repetitive integrations of the neat $\tan'(x)=1+\tan(x)^2$.

A recurrence is obtained there for the $a_n$ coefficients : $$\tag{1}a_n=\begin{cases} &\quad\ 1&n=0\\[6pt] &\frac 1{2n+1} \sum_{k=0}^{n-1} a_k\ a_{n-1-k}&n>0\\ \end{cases}$$

Now let's prove by induction that for $n>0$ : $$\tag{2} a_n<\frac{a_{n-1}}2$$
This is true for $n=1$ since $\displaystyle a_1=\frac 13$ and $a_0=1$
If true for all $n$ from $1$ to $m$ we get from $\;\displaystyle a_{m+1}=\frac 1{2m+3} \sum_{k=0}^{m} a_k\ a_{m-k}\,$ : \begin{align} a_{m+1}&<\frac 1{2m+3}\left(\sum_{k=0}^{m-1} a_k\ \frac{a_{m-1-k}}2+a_m a_0\right)\\ a_{m+1}&<\frac 1{2m+3}\left(\frac{2m+1}2a_m+a_m a_0\right)\\ a_{m+1}&<\frac {a_m}2\\ \end{align} (Martin R. provided first a correct derivation for this part and got my vote)

i.e. the wished conclusion (btw the ration appears to converge to $\dfrac 4{\pi^2}$)

Answer 5

I'll use the notation $c_n$, $n\ge 0$ for the odd coefficients of the Taylor series of the tangent function: $$ \tan(x) = \sum_{n=0}^\infty c_n x^{2n+1} \, . $$ Substituing the Taylor series into the differential equation $$ \frac{d}{dx}\tan(x) = 1 + \tan(x)^2 $$ gives $$ \sum_{n=0}^\infty (2n+1)c_nx^{2n} = 1 + \sum_{n=1}^\infty \left(\sum_{k=0}^{n-1} c_k c_{n-1-k} \right) x^{2n} \, . $$ By comparing the coefficients we get a recursion formula for the $c_n$: $$ \begin{align} c_0 &= 1 \\ c_{n} &= \frac{1}{2n+1} \sum_{k=0}^{n-1} c_k c_{n-1-k} \quad (n \ge 1) \, . \end{align} $$ This can be used to prove $c_{n+1} < c_n$ with (strong) induction. The base case $c_1 = 1/3 < 1 = c_0$ can be verified directly. Now assume that $c_{j+1} < c_j$ holds for $0 \le j \le n-1$. Then $$ \begin{align} c_{n+1} &= \frac{1}{2n+3} \left\{\left(\sum_{k=0}^{n-1} c_k c_{n-k}\right) + c_n c_0 \right\} \\ &< \frac{1}{2n+3} \left\{\left(\sum_{k=0}^{n-1} c_k c_{n-1-k}\right) + c_n \right\} \\ &= \frac{1}{2n+3} \bigl\{(2n+1) c_n + c_n \bigr\} \\ &= \frac{2n+2}{2n+3} c_n < c_n \, . \end{align} $$