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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?

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    $\begingroup$ 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? $\endgroup$ – Sangchul Lee Mar 6 at 17:39
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    $\begingroup$ 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 $K$ ? $\endgroup$ – Claude Leibovici Mar 7 at 4:31
  • $\begingroup$ Sorry for the late response @ClaudeLeibovici . Yes, I would like to compute $K$, which would be $$\sum_{k=2}^{\infty}{\frac{|B_k|}{k!}}$$ $\endgroup$ – Gnumbertester Mar 9 at 17:45
  • $\begingroup$ @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. $\endgroup$ – Gnumbertester Mar 9 at 17:47
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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.

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    $\begingroup$ You can use the fact that $B_n$ alternate for $n > 1$, so $|B_{2n}| = (-1)^n B_{2n}$ to show your $.084756 \ldots = i/ (1- e^i) - i/2 + 1$. $\endgroup$ – Jair Taylor Mar 10 at 8:20
  • $\begingroup$ @JairTaylor. Shame on me ! $\endgroup$ – Claude Leibovici Mar 10 at 8:28
  • $\begingroup$ Thank you @ClaudeLeibovici! This certainly helps. I will look into those approximations. $\endgroup$ – Gnumbertester Mar 10 at 14:35
  • $\begingroup$ @JairTaylor. I didn't recognize that, thank you! $\endgroup$ – Gnumbertester Mar 10 at 14:36
  • $\begingroup$ @ClaudeLeibovici No shaming necessary! $\endgroup$ – Jair Taylor Mar 10 at 17:08

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