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I am trying to find a closed form expression of the following sum in terms of $n$ (if it exists) where $B_{k}$ is the $k$th Bernoulli number.

$$\sum_{k=2}^{\infty}{\frac{|{B_{k}|}}{k!}(\cos(n)-1)}$$

The problem I am having is that since the summation function isn't in terms of elementary functions, I cannot (or do not know how to) evaluate it using traditional means.

I am aware of the Bernoulli number generating function which is given by the following, however I do not know how to apply it here, especially with the absolute value. $$\frac{t}{e^t-1}=\sum_{k=1}^{\infty} B_k \frac{t^k}{k!}$$

Here is my attempt:

Write out the first few terms.

$$\frac{1}{12}(\cos{(n)}-1)+0+\frac{1}{720}(\cos(n)-1)+0+\frac{1}{30240}(\cos(n)-1)+...$$

Let $w=\cos{(n)}$, then we have the following: $$\frac{1}{12}w-\frac{1}{12}+0+\frac{1}{720}w-\frac{1}{720}+0+\frac{1}{30240}w-\frac{1}{30240}+...$$

I can tell that $-\frac{1}{12}-\frac{1}{720}-\frac{1}{30240}-...$ converges, however, I do not know how to write this in a closed form because it is a series of Bernoulli numbers divided by factorials. I also do not know how to proceed with the elements in terms of $w$.

Any help with finding a closed form expression of this series would be appreciated. Does it not have a closed form equivalent?

rtybase
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Gnumbertester
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    Given that the signs of Bernoulli numbers for even indices alternate, how about replacing $t$ by $ie^{ix}$ in the exponential generating function that you are mentioning in the posting? – Sangchul Lee Mar 06 '19 at 17:39
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    I am sorry but I do not understand what is the problem. Since $n$ does not depend on $k$, then $$\sum_{k=2}^{\infty}{\frac{|{B_{k}|}}{k!}}=(\cos(n)-1)\sum_{k=2}^{\infty}{\frac{|{B_{k}|}}{k!}}=K(\cos(n)-1)$$ Do you want to compute $K$ ? – Claude Leibovici Mar 07 '19 at 04:31
  • Sorry for the late response @ClaudeLeibovici . Yes, I would like to compute $K$, which would be $$\sum_{k=2}^{\infty}{\frac{|B_k|}{k!}}$$ – Gnumbertester Mar 09 '19 at 17:45
  • @SangchulLee I do not know how to apply that yet. The function I mentioned in the OP is slightly different from the sum I am trying to compute. – Gnumbertester Mar 09 '19 at 17:47

2 Answers2

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Not an answer but too long for a comment.

I did not find any way to formulate the result but numerical calculations of $$S_p=\sum_{k=2}^{2^p}{\frac{|B_{k}|}{k!}}$$ show that it converges very quickly $$\left( \begin{array}{cc} p & S_{p} \\ 1 & \color{red} {0.08}33333333333333333333333333333333333333333333333333333333333 \\ 2 & \color{red} {0.0847}222222222222222222222222222222222222222222222222222222222 \\ 3 & \color{red} {0.0847561}177248677248677248677248677248677248677248677248677249 \\ 4 & \color{red} {0.0847561391437}652311648901025392362668013669154866374494966673 \\ 5 & \color{red} {0.0847561391437740403659902}790226013248134712189289954349942019 \\ 6 & \color{red} {0.0847561391437740403659902805155916881209460259919329}549322065 \\ 7 & \color{red} {0.0847561391437740403659902805155916881209460259919329978167920} \\ 8 & \color{red} {0.0847561391437740403659902805155916881209460259919329978167920} \end{array} \right)$$ In fact, to be more precise $S_{8}-S_{7}\approx 3.54\times 10^{-104}$. Inverse symbolic calculators do not identify the number $0.084756139143774\cdots$.

May be, you could use the simplest asymptotics $$|B_{2k}|\sim 4\sqrt{\pi\,k}\,\left(\frac k{\pi\,e}\right)^{2k} $$ as well as the simplest form of Stirling approximation for the factorial to get an idea of the remainder.

Claude Leibovici
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Yes a closed form exists!

An entire working process with dead ends included below:

First we recall that The Even Index Bernoulli Numbers Alternate in Signs

So our goal is to somehow remove the absolute value with a suitable method of alternating signs.

A first approach is to observe:

$$ \sum_{k=2}^{\infty} \frac{|B_k|}{k!} = \sum_{k=2}^{\infty} (-1)^{\frac{n(n-1)}{2}+1}\frac{B_k }{k!}$$

That quadratic exponent in the $(-1)$ strongly suggests there is something we can do with theta functions here. But theta functions are hard and complicated so before jumping off into the deep end we try another idea

Recalling that

$$ \sum_{k=0}^{\infty} \frac{B_k}{k!}x^n = \frac{x}{e^x-1} $$

(You have a typo where your indexing begins from $1$ )

We then have:

$$ \sum_{k=2}^{\infty} \frac{B_k}{k!}x^n = \frac{x}{e^x-1} -1 + \frac{x}{2} = g(x)$$

Now with the help of wolfram alpha we see the first few terms here are

$$g(x) = \frac{|B_2|}{2!}x^2 - \frac{|B_4|}{4!}x^4 + \frac{|B_6|}{6!}x^6 - ... $$

If we let $x = iv$ then we find:

$$g(iv) = -\frac{|B_2|}{2!}v^2 - \frac{|B_4|}{4!}v^4 - \frac{|B_6|}{6!}v^6 - ... $$

Therefore:

$$-g(iv) = \sum_{k=2}^{\infty} \frac{|B_k|}{k!}v^n = -\frac{iv}{e^{iv} -1} +1 - \frac{iv}{2}$$

Recall you were seeking: $(\cos(n)-1) \sum_{k=2}^{\infty} \frac{|B_k|}{k!} $ therefore we set $v=1$ to find a closed form for our magic constant:

$$ \sum_{k=2}^{\infty} \frac{|B_k|}{k!} = -g(i)= -\frac{i}{e^i - 1} + 1 - \frac{i}{2}$$

We confirm with Wolfram Alpha that this equal to $0.084756...$ which numerically agrees with Claude Leibovicci's Result.

A now deleted answer by E.Dioekema correctly removes the complex exponentials by expressing the answer as $\frac{2 - \cot \left( \frac{1}{2} \right)}{2}$. A good exercise is to use Euler's theorem of complex exponentials to trig functions and verify that E.Dioekema's result is indeed correct.

Sidharth Ghoshal
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