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The below series is convergent series by the ratio test but i'm no able to know if this series have a positive sum , and i don't succeed to check if it has a closed form ,Then my question here is :

Question: Is this : $\displaystyle\sum_{n=1}^{+\infty}\frac{(-1)^n \log n}{n!}$ a positive sum ?

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    $\begingroup$ It is fast convergent series, so it is enough to estimate few first terms. $\endgroup$ – Oleg567 Dec 5 '17 at 15:59
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    $\begingroup$ The alternating series test comes with an estimate for the error. That should be enough to establish the sign of the sum. $\endgroup$ – Giuseppe Negro Dec 5 '17 at 16:02
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For $n \geq 2$, the expression $\frac{\log n}{n!}$ is strictly decreasing, as $\frac{\log (n+1)}{(n+1)!} < \frac{\log n}{n!} \iff \log (n+1) < (n+1) \log n \iff \log_n (n+1) < n+1$, which is true, because $\log_n (n+1) < \log_n (n^2) = 2$. Therefore, $$\sum_{n=1}^\infty \frac{(-1)^n \log n}{n!}$$ is an alternating series with terms that strictly decrease in magnitude after the first nonzero term, so the series has the same sign as the first nonzero term, which is positive.

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-Teacher, what is the best way for dealing with series with oscillating terms?
-To invoke the Laplace transform, of course.

Since $\log(n)=\int_{0}^{+\infty}\frac{e^{-x}-e^{-nx}}{x}\,dx $ by Frullani's theorem, by multiplying both sides by $\frac{(-1)^n}{n!}$ and summing over $n\geq 1$ we have: $$ \sum_{n\geq 1}\frac{(-1)^n \log(n)}{n!} = \int_{0}^{+\infty}\frac{\color{blue}{1+e^{-x-1}-e^{-x}-e^{-e^{-x}}}}{x}\,dx $$ where it is tedious, but fairly simple, to check that the blue term is positive for any $x>0$.
It follows that the LHS is positive as well. Additionally, since the integrand function in the RHS behaves like $\left(1-\frac{2}{e}\right)\exp\left[-\frac{(e-1)}{2(e-2)}x\right]$ in a right neighbourhood of the origin, the value of the LHS is approximately $\frac{2(e-2)^2}{e(e-1)}\approx 0.22$.


Actually this trick works best for checking the sign of slowly convergent, oscillating series. Here the term $\frac{\log(n)}{n!}$ goes to zero incredibly fast, so it is not really needed, as already shown by Connor Harris above.

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