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Continuing my search of an integral expression for the sequence:

$$a_s=\sum_{n=0}^s {s \choose n} \frac{B_n}{n!}$$

(Where $B_n$ is the $n$th Bernoulli Number, $B_1$ is taken to be $\frac{1}{2}$ and $s$ must be a positive integer, see my other post) I came across the following development, which may be incorrect, since there is numerical evidence to think so. I would be interested in finding the flaw and trying to correct it (if possible) to get a correct result.

Let's start by pointing out the fact that, by the definition of binomial coefficient:

$$\sum_{n=0}^s {s \choose n} \frac{B_n}{n!}=\sum_{n=0}^\infty {s \choose n} \frac{B_n}{n!}$$

Then, since:

$${s \choose n}= \frac{1}{2\pi} \int_{-π}^π e^{itn}(1+e^{it})^s \, dt$$

We have that:

$$\frac{1}{2\pi} \sum_{n=0}^\infty \int_{-π}^π e^{itn}(1+e^{it})^s \, dt \frac{B_n}{n!}$$

Up to here, everything seems to work fine.

Then, as the sum is absolutely convergent for positive integer $s$, say

$$\frac{1}{2\pi} \sum_{n=0}^\infty \left | \int_{-π}^π \frac{B_n}{n!} e^{itn}(1+e^{it})^s \, dt \right | <\infty$$

We are allowed to interchange the sum and the integral operators (by the Fubini/Tonelli theorems, see this other question) to get:

$$\frac{1}{2\pi} \int_{-π}^π (1+e^{it})^s \sum_{n=0}^\infty e^{itn} \frac{B_n}{n!} \, dt$$

By the Generating Function of Bernoulli Numbers, and since for any complex $z$ and integer $n$, $e^{zn}=(e^z)^n$,

$$\sum_{n=0}^\infty e^{itn} \frac{B_n}{n!}=\frac{e^{it}}{1-e^{-e^{it}}}$$

So that our original function/sequence can be expressed as:

$$\frac{1}{2\pi} \int_{-π}^π (1+e^{it})^s \frac{e^{it}}{1-e^{-e^{it}}} \, dt$$

The problem is that, fom example, setting $s=3$ in our original sequence (the sum), we would get $a_3=\frac{11}{4}$, but the integral will equal $1$. This error continues for all other values of $s$ computed.

Any help with this? Where is the flaw and how could it be solved?

Thank you very much.

user3141592
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  • Have you numerically checked each step? That is always a good way to go when you end up with a weird numerical answer. Find the step where numerics start to go astray – Brevan Ellefsen Aug 26 '17 at 18:08
  • @Brevan I have checked up to the sentence "Up to here, everything seems to work fine" by using Wolfram alpha. From there on, the only step I can check is the last one, which gives an erroneous result – user3141592 Aug 26 '17 at 18:10
  • I would also want to know what is the precise error there, to understand the concept behind it and avoid future similar errors – user3141592 Aug 26 '17 at 18:19
  • I'll check this out later in Mathematica or something. I usually use simpler numeric tools, but none I generally use have the Bernoulli Numbers :/ – Brevan Ellefsen Aug 26 '17 at 23:10
  • I do not know about Mathematica, since I do not own that program, but at least in Wolfram Alpha the command is called NorlundB – user3141592 Aug 26 '17 at 23:22
  • ?? At least in English, Wolfram Alpha commands are the same as they would be in Mathematica (though wolfram Alpha has far less). In Mathematica this would be the command BernoulliB – Brevan Ellefsen Aug 26 '17 at 23:30
  • Are you using NorlundB in order to get $B_1 = 1/2$? I would think NorlundB still uses $B_1 = -1/2$... That's what the documentation seems to say at least – Brevan Ellefsen Aug 26 '17 at 23:38
  • NorlundB gives you the Bernoulli polynomial, so that $B_n$ would be NorlundB[n,1,1] – user3141592 Aug 26 '17 at 23:43
  • Ok. Be aware that Mathematica has $B_1 = -1/2$ though, which seems to contradict your post – Brevan Ellefsen Aug 26 '17 at 23:58
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    Your definition for the Binomial Coefficient is incorrect, it should be $e^{-itn}$, see here for instance. – Paul LeVan Aug 28 '17 at 01:03
  • @Paul but, since $n$ is an integer, can't we say in this case that $e^{int}=e^{itn}=(e^{it})^n$? – user3141592 Aug 28 '17 at 01:09
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    That is a true statement, but, in the integral representation of the Binomial Coefficient, the $e^{itn}$ should be $e^{-itn}$, which are not equivalent. – Paul LeVan Aug 28 '17 at 01:14
  • Oh God, the minus sign. Thanks, I will correct it and repair my post as soon as possible – user3141592 Aug 28 '17 at 01:15
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    There does not seem to be any error here – Paul LeVan Aug 28 '17 at 01:29
  • I have re-re-checked my original notes and you were right, at the middle of that mess I missed the $-$ sign. Thank you very much and sorry for my stupid mistake. – user3141592 Aug 28 '17 at 01:29
  • I have deleted the missign signs from the question so that you can add an answer – user3141592 Aug 28 '17 at 01:31

1 Answers1

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As stated in the comments, the correct integral representation for the binomial coefficients is given by: $$ {s \choose n}= \frac{1}{2\pi} \int_{-π}^π e^{-itn}(1+e^{it})^s \, dt $$ With this change, all of the results are accurate.

Paul LeVan
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