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The Euler-Maclaurin formula says (from Concrete Mathematics section 9.5)

\[ \sum_{a\le{}k< b}f(k)=\int_a^bf(x)dx+\left.\sum_{k=1}^m\frac{B_k}{k!}f^{(k-1)}(x)\right|_a^b+R_m \] where $\displaystyle{}R_m=(-1)^{m+1}\int_a^b\frac{B_m(\{x\})}{m!}f^{(m)}(x)dx$, integers $a\le{}b$ and $m\ge1$.

However, let $a=0$, $b=n$, and consider the series \[ \int_0^nf(x)dx+\sum_{k=1}^\infty\frac{B_k}{k!}\left(f^{(k-1)}(n)-f^{(k-1)}(0)\right) \] Unfortuntely, the series usually diverges. So, I wonder whether there is some sufficient condition (easy to verify) that the series converges?

Especially when the series $\sum_{k=0}^\infty{}f(k)$ converges, or more stronger, converges absolutely, what about those conditions?

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1  
It diverges, yes, but the behavior of the Euler-Maclaurin series is usually asymptotic; you can truncate the series up to some point, and that truncation is a good approximation. – J. M. Jun 9 '12 at 14:08
    
@J.M. Yes, and O-manipulation usually works well, but I wonder when it converges. – Frank Science Jun 10 '12 at 2:12
up vote 8 down vote accepted

We have the estimate $$ \lim_{k\to\infty}\left(\frac{\left|B_{2k}\right|}{(2k)!}\right)^{\Large{\frac{1}{2k}}}=\frac{1}{2\pi}\tag{1} $$ Thus, for the series $$ \sum_{k=1}^\infty\frac{B_{2k}}{(2k)!}\left(f^{(2k-1)}(n)-f^{(2k-1)}(0)\right)\tag{2} $$ to converge, we need $$ \lim_{k\to\infty}\left(f^{(2k-1)}(n)-f^{(2k-1)}(0)\right)^{\Large{\frac{1}{2k}}}\le2\pi\tag{3} $$ Since $R$, the radius of convergence of $f$, satisfies $$ \frac1R=\limsup_{k\to\infty}\left(\frac{f^{(k)}(x)}{k!}\right)^{\Large{\frac{1}{k}}}\tag{4} $$ $(3)$ and $(4)$ would seem to require that $\dfrac1R=0$; that is, $f$ be entire.

In fact, according to this result, if the Fourier Transform of $f$ is contained in $[-1,1]$, then $$ \limsup_{k\to\infty}\left(f^{(k)}\right)^{\Large{\frac{1}{k}}}\le2\pi\tag{5} $$ and therefore, the series in $(2)$ converges.

Of course, the Euler-Maclaurin Sum Formula is finite on polynomials, and so the series in $(2)$ also converges for polynomials.

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I believe you mean absolute value on your fist equation since $B_{2n}$ is an oscillating funtion with both positve and negative values. – Herman Jaramillo 19 hours ago
    
In addition to the sign, I believe that you are missing also a $2^{\frac{1}{2k}}$ on the first equation. Please check the following: en.wikipedia.org/wiki/Bernoulli_number#Asymptotic_approximation link. Oh, but $\lim_{k \to \infty} 2^{frac{1}{2k}$ is 1. So that part is fine. – Herman Jaramillo 17 hours ago

From Wikipedia

\begin{eqnarray*} B_{2 k} &=& (-1)^{k+1} \frac{2 \, (2 \, k)!}{ (2 \, \pi)^{2 k}} \sum_{n=1}^{\infty} \frac{1}{n^{2k}} = (-1)^{k+1} \frac{2 \, (2 \, k)!}{ (2 \, \pi)^{2 k}} + \mathcal{O} \left ( \left ( \frac{1}{2} \right)^{2k} \right ) \end{eqnarray*}

Then:

\begin{eqnarray} \lim_{k \to \infty} \left | \frac{B_{2k}}{2 \ (2k)!} \right |^{\frac{1}{2k}} = \frac{1}{2 \pi} \label{toseries}. \end{eqnarray}

Since $\lim_{k \to \infty} (1/2)^{1/2k}=1$ we can also write

\begin{eqnarray} \lim_{k \to \infty} \left | \frac{B_{2k}}{(2k)!} \right |^{\frac{1}{2k}} = \frac{1}{2 \pi} \label{toseries2}. \end{eqnarray}

Let us call the $k$ term of the series:

\begin{eqnarray*} a_k = \left . \frac{B_{2k}}{(2k)!} f^{(2k-1)}(x) \; \right |_{x=a}^b \end{eqnarray*}

then the radius of convergence of the series is found from the limit $\lim_{k \to \infty} |a_{k+1}|/ |a_{k}|$. In this case we have

\begin{eqnarray*} R = \lim_{k \to \infty} \left | \frac{ \left . B_{2k +2} f^{(2k+1)}(x) \; \right |_{x=a}^b (2k)!} {\left . B_{2k} f^{(2k-1)} \right |_{x=a}^b (2k+2)!} \right | = \lim_{k \to \infty} \left | \frac{ \left .f^{(2k+1)} \; \right |_{x=a}^b} {\left . f^{(2k-1)} \;\right |_{x=a}^b} \right |. \end{eqnarray*}

We need $R < 1$ for convergence.

That is, we need to have the condition:

\begin{eqnarray} \lim_{k \to \infty} \left | \frac{ \left .f^{(2k+1)} \; \right |_{x=a}^b} {\left . f^{(2k-1)} \;\right |_{x=a}^b} \right | < 1 \end{eqnarray}

for convergence of the series.

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