$\sum_{n=2}^{\infty} (-1)^n \frac{1}{\log n}$ =?

Question

The alternating series, $s_n=\sum_{i=2}^{n} (-1)^i \frac{1}{\ln i}$, (with $\ln$ being the natural logarithm) converges for $n\to\infty$ which can be seen e.g. by the Leibniz test. Can the limit be expressed in a finite closed expression of otherwise known constants?

I got interested while thinking about the values of the sine integral at integer multiples of $\pi$ where I suspect some relation.

Answer 1

Doing my usual pairing of even and odd, if $s_n =\sum_{i=2}^{n} (-1)^i \dfrac{1}{\ln i} $ then

$\begin{array}\\ s_{2n+1} &=\sum_{i=2}^{2n+1} (-1)^i \dfrac{1}{\ln i}\\ &=\sum_{i=1}^{n} (\dfrac{1}{\ln (2i)}-\dfrac{1}{\ln (2i+1)})\\ &=\sum_{i=1}^{n} \dfrac{\ln (2i+1)-\ln (2i)}{\ln (2i)\ln (2i+1)}\\ &=\sum_{i=1}^{n} \dfrac{\ln (1+1/2i)}{\ln (2i)\ln (2i+1)}\\ &<\sum_{i=1}^{n} \dfrac{1/2i}{\ln (2i)\ln (2i+1)}\\ &=\sum_{i=1}^{n} \dfrac{1}{2i\ln (2i)\ln (2i+1)}\\ \end{array} $

and this converges by comparison with $\sum \dfrac1{i \ln^2 i} $.

However, this does converge quite slowly. The integral test, using $\int_2^n \dfrac{dt}{t\ln^2(t)} =\dfrac1{\ln(2)}-\dfrac1{\ln(n)} $ shows how slowly. This, of course, agrees with the result that the sum of an alternating series is between any two consecutive partial sums.

For example, if $n = 10^{10}$, the error is about $\dfrac1{2\ln(10^{10})} =\dfrac1{20\ln(10)} \gt .01 $.

If we apply Euler's transform (https://en.wikipedia.org/wiki/Series_acceleration),

$\begin{array}\\ s &=\sum_{n=0}^{\infty} (-1)^n \dfrac{1}{\ln (n+2)}\\ &=\sum_{n=0}^{\infty} \dfrac{(-1)^n}{2^{n+1}}\sum_{k=0}^n(-1)^k\binom{n}{k} \dfrac{1}{\ln (n-k+2)}\\ &\approx\sum_{n=0}^{\infty} \dfrac{(-1)^n}{2^{n+1}}(\dfrac1{\ln(x+2)})^{(n)}|_{x=2}\\ &\approx\sum_{n=0}^{\infty} \dfrac{(-1)^n}{2^{n+1}}((-1)^{n-1}\dfrac{(n-1)!}{(x+2)^n\ln(x+2)})|_{x=2}\\ &=-\sum_{n=0}^{\infty} \dfrac{(n-1)!}{2^{3n+1}\ln(4)}\\ \end{array} $

and this actually diverges! (modulo any errors on my part)

That's enough for now.

Answer 2

Another comment, with no closed form at the end.

Start with the polylogarithm $$ \sum_{n=1}^\infty a^n n^{-s} = \mathrm{Li}_s(a) $$ So that $$ \sum_{n=1}^\infty (-1)^n n^{-s} = \mathrm{Li}_s(-1) \\ \sum_{n=2}^\infty (-1)^n n^{-s} = 1 + \mathrm{Li}_s(-1) $$ Integrate with respect to $s$ from $0$ to $+\infty$ $$ \sum_{n=2}^\infty (-1)^n \frac{1}{\log(n)} = \int_0^\infty\big(1+\mathrm{Li}_s(-1)\big)ds \approx 0.92429989722293885595957018136 $$ Unfortunately, this also has no known closed form as far as I can find.

Answer 3

This value can be computed using the Euler-Maclaurin Sum Formula.

We will start with $$ \begin{align} g_1(n) &=\sum_{k=2}^n\frac1{\log(k)}\\ &=\,\scriptsize C_1+\operatorname{li}(n)+\frac1{2\log(n)}-\frac1{12n\log(n)^2}+\frac1{360n^3\log(n)^2}\left(1+\frac3{\log(n)}+\frac3{\log(n)^2}\right)\\ &\,\scriptsize -\,\frac1{15120n^5\log(n)^2}\left(12+\frac{50}{\log(n)}+\frac{105}{\log(n)^2}+\frac{120}{\log(n)^3}+\frac{60}{\log(n)^4}\right)\\[3pt] &\,\scriptsize +\,O\!\left(\frac1{n^7\log(n)^2}\right)\tag1 \end{align} $$ where $\operatorname{li}(n)$ is the Logarithmic Integral Function, and $$ \begin{align} g_2(n) &=\sum_{k=1}^n\frac1{\log(2k)}\\ &\,\scriptsize =C_2+\frac12\operatorname{li}(2n)+\frac1{2\log(2n)}-\frac1{12n\log(2n)^2}+\frac1{360n^3\log(2n)^2}\left(1+\frac3{\log(2n)}+\frac3{\log(2n)^2}\right)\\ &\,\scriptsize -\,\frac1{15120n^5\log(2n)^2}\left(12+\frac{50}{\log(2n)}+\frac{105}{\log(2n)^2}+\frac{120}{\log(2n)^3}+\frac{60}{\log(2n)^4}\right)\\[3pt] &\,\scriptsize +\,O\!\left(\frac1{n^7\log(n)^2}\right)\tag2 \end{align} $$ To get the alternating series, we compute $$ \begin{align} \sum_{k=2}^{2n}\frac{(-1)^k}{\log(k)} &=2\sum_{k=1}^n\frac1{\log(2k)}-\sum_{k=2}^{2n}\frac1{\log(k)}\\ &=2g_2(n)-g_1(2n)\\[6pt] &\,\scriptsize =2C_2-C_1+\frac1{2\log(2n)}-\frac1{8n\log(2n)^2}+\frac1{192n^3\log(2n)^2}\left(1+\frac3{\log(2n)}+\frac3{\log(2n)^2}\right)\\ &\,\scriptsize -\,\frac1{7680n^5\log(2n)^2}\left(12+\frac{50}{\log(2n)}+\frac{105}{\log(2n)^2}+\frac{120}{\log(2n)^3}+\frac{60}{\log(2n)^4}\right)\\[3pt] &\,\scriptsize +\,O\!\left(\frac1{n^7\log(n)^2}\right)\tag3 \end{align} $$ Note that the $\operatorname{li}(n)$ terms cancel.

We can use $(3)$ with $n=1000000$, or we can extend $(3)$ to contain all terms bigger than $O\!\left(\frac1{n^{11}}\right)$ and use $n=10000$. In either case, we get

$$ \begin{align} \sum_{k=2}^\infty\frac{(-1)^k}{\log(k)} &=2C_2-C_1\\ &=0.92429989722293885595957018135959005377331939789\tag4 \end{align} $$