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Say, we use the definition:

Bernoulli numbers arise in Taylor series in the expansion $$\frac{x}{e^x-1}=\sum_{k=0}^\infty B_k \frac{x^k}{k!}$$

and then derive power series representations of the functions:

\begin{align} \\ x \coth(x) &= \sum_{n=0}^\infty \frac{2^{2n}}{(2n)!}B_{2n}x^{2n} \\ x \cot(x) &= \sum_{n=0}^\infty (-1)^n\frac{2^{2n}}{(2n)!}B_{2n}x^{2n} \\ \tan(x) &= \sum_{n=1}^\infty (-1)^{n-1}\frac{2^{2n}(2^{2n}-1)}{(2n)!}B_{2n}x^{2n-1} \end{align}

If we take for granted that $\frac{x}{e^x-1}$ is analytic in $x_0=0$, can we conclude the following?

  1. $x \coth(x) = \frac{x}{2} + \frac{x}{e^x-1}$, therefore $x \coth(x)$ is also analytic in $x_0=0$ (as a sum of analytic functions).

  2. $x \cot(x) = (ix)\coth(ix)$, therefore $x \cot(x)$ is also analytic in $x_0=0$ (as a composition of analytic functions).

  3. $\tan(x) = \cot(x) - 2\cot(2x)$, therefore $x \tan(x)$ is also analytic in $x_0=0$ (as a sum of analytic functions).

This (if correct) gives us that the convergence radiuses are greater than zero.

But what else can be said about their radiuses of convergence? How can we calculate them? The Bernoulli numbers scare me away from applying the usual root/ratio tests.

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The identity (1) should be $\frac{x}{2} \coth(\frac{x}{2}) = \frac{x}{2} + \frac{x}{e^x - 1}$. Consequently the singularities for $x \coth(x)$ and $x \cot(x)$ are only half the distance from the origin, i.e. $\pi$ away. A little care is needed in the argument for (3), because $\cot(x)$ is singular at the origin, although $\tan(x)$ is analytic in a neighborhood. –  hardmath May 7 '13 at 18:38

1 Answer 1

up vote 1 down vote accepted

The radius of convergence of the first shown power series centered at the origin is the distance to the nearest singularity in the complex plane, namely at $\pm 2\pi i$. So radius of convergence is $2 \pi$.

Note that the denominator of $x/(e^x -1)$ gives a simple pole at every nonzero integer multiple of $2 \pi i$ owing to periodicity of the exponential function.


We'd arrive at the same radius of convergence by applying the ratio test, using an asymptotic approximation for the Bernoulli numbers:

$$ |B_{2k}| \approx 4 \sqrt{\pi k} (\frac{k}{\pi e})^{2k} $$

Since all the odd index Bernoulli numbers after the first are zero:

$$ \sum_{k=0}^\infty \frac{B_k x^k}{k!} = \frac{-x}{2} + \sum_{k=0}^\infty \frac{B_{2k} x^{2k}}{(2k)!} $$

we should apply the ratio test to consecutive even (nonzero) terms of the series:

$$ \lim_{k \rightarrow \infty} |\frac{B_{2k+2}}{B_{2k}} \frac{x^2}{(2k+2)(2k+1)}| \lt 1 $$

Replacing $B_{2k},B_{2k+2}$ by the asymptotic approximations gives:

$$ |\frac{B_{2k+2}}{B_{2k}}| \approx \sqrt{1+\frac{1}{k}} (1+\frac{1}{k})^{2k} \frac{(k+1)^2}{(\pi e)^2} $$

Combining this approximation with the rest of the ratio test expression and taking the limit leads to:

$$ \lim_{k \rightarrow \infty} \sqrt{1+\frac{1}{k}} (1+\frac{1}{k})^{2k} \frac{(k+1)^2}{(2k+2)(2k+1)} \frac{x^2}{(\pi e)^2} = e^2 \cdot \frac{1}{4} \cdot \frac{x^2}{\pi^2 e^2} = \frac{x^2}{4\pi^2} $$

This ratio test limit is less than one precisely when $|x| \lt 2\pi$, and that again gives the radius of convergence as $2\pi$.

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We didn't have the stuff (singularity, simple pole) in the course yet. It looks like there is no "elementary" way to get more information about the radiuses (except the positivity via analyticity). Thank you for the answer! –  Leo Schmidt May 7 '13 at 18:03
1  
@mathusiast: If you wanted to try a ratio test on this question, have a look at the (even) Bernoulli numbers' asymptotic approximation $|B_{2n}| \approx 4 \sqrt{\pi n}(\frac{n}{\pi e})^{2n}$. More rigorous bounds are described here. –  hardmath May 7 '13 at 18:21

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