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I wonder what is the sum of this series?

$$\sum_{n=1}^\infty\frac{(-1)^{n-1} B_n}{n}$$

where $B_n$ are Bernoulli numbers. Wolfram Alpha does not help.

P.S. As this series diverges I am interested in generalized summation. Mathematica fails to find the sum using Abel, Borel, Dirichlet, Cesaro and Euler's regularizations.

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  • $\begingroup$ i would have guessed that borel should work (because it essentially kills the $n!$ in the asymptotic expansion)...have you tried it by hand? Mathematica is no always trustable in summing up divergent series... $\endgroup$ – tired Sep 9 '15 at 16:43
  • $\begingroup$ @tired I do not know how to use it to obtain closed form. I do not need a numerical result... $\endgroup$ – Anixx Sep 9 '15 at 16:44
  • $\begingroup$ i have no time today to get into details, but my first starting point would be to represent the Bernoulli numbers by a contour integral and then trying the borel machine on this... $\endgroup$ – tired Sep 9 '15 at 16:48
  • $\begingroup$ @tired I suspect the answer is $-2\gamma$. Need to be verified $\endgroup$ – Anixx Sep 9 '15 at 19:18
  • $\begingroup$ @have you used my suggestion? then i'm confident that the euler constant will show up somewhere! $\endgroup$ – tired Sep 9 '15 at 19:47
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With a purely formal manipulation, $$\frac{1}{e^t-1}-\frac{1}{t}=\sum_{n\geq 1}\frac{B_n}{n!}t^{n-1} \tag{1}$$ leads to: $$\frac{1}{e^{-t}-1}+\frac{1}{t}=\sum_{n\geq 1}\frac{(-1)^{n-1} B_n}{n!}t^{n-1}\tag{2}$$ hence by multiplying both sides by $e^{-t}$ and integrating them over $\mathbb{R}^+$, $$\int_{0}^{+\infty}\left(\frac{1}{t e^t}-\frac{1}{e^t-1}\right)\,dt = \sum_{n\geq 1}\frac{(-1)^{n-1} B_n}{n}\tag{3}$$ but the LHS of $(3)$ equals $\color{red}{-\gamma}$.

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    $\begingroup$ nice, borel summation in action (+1) $\endgroup$ – tired Sep 9 '15 at 20:40
  • $\begingroup$ This well can be the case! $\endgroup$ – Anixx Sep 9 '15 at 20:40
  • $\begingroup$ By the way, why you have $\frac{1}{e^t-1}-\frac{1}{t}$ rather than simply $\frac{1}{e^t-1}$ in the first identity? Generating function is $\frac{1}{e^t-1}$. $\endgroup$ – Anixx Sep 9 '15 at 20:59
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    $\begingroup$ @Anixx: the sum on the right starts at $n=1$, so we have to subtract a term. $\endgroup$ – Jack D'Aurizio Sep 9 '15 at 21:01
  • $\begingroup$ Ah I see, thanks! $\endgroup$ – Anixx Sep 9 '15 at 21:03
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According to this source, the Bernoulli numbers satisfy the inequality $|B_{2n}| \ge \frac{(2n)!}{(2\pi)^{2n}}$. This means the sequnce $\frac{B_n}{n}$ is unbounded and the sum doesn't converge.

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The sum doesn't converge because the general term $B_n/n$ goes to $+\infty$ in absolute value, by the asymptotic approximation here.

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  • $\begingroup$ The series has alternating sign. $\endgroup$ – Anixx Sep 9 '15 at 16:05
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    $\begingroup$ @Anixx: The absolute value of the terms does not tend to $0$. $\endgroup$ – Clayton Sep 9 '15 at 16:06
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    $\begingroup$ @Anixx: The most basic test for divergence says that if $|a_n|\not\to0$, then $\sum a_n$ cannot converge. $\endgroup$ – Clayton Sep 9 '15 at 16:07
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    $\begingroup$ @Anixx So does the series $\sum (-1)^n n$. Alternating sign doesn't get you anything if the terms don't go to zero in absolute value. $\endgroup$ – Ben Sheller Sep 9 '15 at 16:08
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    $\begingroup$ @Anixx It may alternate, but does not go to zero. Regardless of the sign, the general term of the series must converge to zero for the series to converge: it is a necessary (not sufficient) condition. $\endgroup$ – Clement C. Sep 9 '15 at 16:08

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