What is $\sum_{n=0}^\infty\frac{(-1)^n}{\log^* (n+2)}$? I'm interested in approximating $\sum_{n=0}^\infty\frac{(-1)^n}{\log^* (n+2)}$.
Here, $\log^*$ is the iterated log function.
I tried several solvers (e.g., Wolfram), but none seem to yield a solution.
Any ideas?
 A: A function $\log^*$ is monotone and $\sum_{n=0}^\infty\frac{(-1)^n}{\log^* (n+2)}$ alternating series. Let $L$ denote the sum of the series, then the partial sum approximates $L$ with error bounded by the next omitted term: see, e.g. https://en.wikipedia.org/wiki/Alternating_series_test
Hence $L_N = \sum_{n=0}^N \frac{(-1)^n}{\log^* (n+2)}$ is a good approximation when $\frac{1}{\log^* (N+3)}$ is small enough. Unfortunately $\log^*$ grows too slowly. E.g. $\frac{1}{\log^* (N+3)} < \frac1{10}$ for $N > \exp( \exp ( \ldots (\exp (e)) \ldots )$. Fortunately, for "many" values of $n$ we have $a_n = -a_{n+1}$, where $a_n = \frac{(-1)^n}{\log^* (n+2)}$ and it makes summation much easier.
A: The sum is in [0.5, 0.7], and probably nobody will ever be able to say anything further.
Here are some details about @BotnakovN's idea (that repeated values of $\log^*$ make summation easier). Let
\begin{align}
a_0 &= 1 \\
a_1 &= \lfloor e \rfloor = 2\\
a_2 &= \lfloor e^e \rfloor = 15\\
a_3 &= \lfloor e^{e^e} \rfloor = 3814279\\
a_4 &= \lfloor e^{e^{e^e}} \rfloor = 233\dots\text{(many digits)}\dots021\\
& \vdots
\end{align}
(so $a_k$ is a tower of $k$ copies of $e$ rounded down to the nearest integer, https://oeis.org/A056072; see https://oeis.org/A085667 and links for $a_4$). Then
$$\log^* n = k \iff a_{k-1} < n < a_k,$$
so
$$ \sum_{n=2}^\infty \frac{(-1)^n}{\log^* n} = \sum_{k=1}^\infty \frac1k \sum_{n=a_{k-1}+1}^{a_k} (-1)^n = \sum_{k=1}^\infty \frac{s_k}{k},$$
where $s_1 = 1$ and, for $k > 1$,
$$
s_k = \begin{cases}
0 &: \text{if}~a_{k-1} \equiv a_k \pmod 2,\\
1 &: \text{if}~(a_{k-1}, a_k) \equiv (1, 0) \pmod 2,\\
-1 &: \text{if}~(a_{k-1}, a_k) \equiv (0, 1) \pmod 2.
\end{cases}
$$
Put another way, let $i_1, i_2, \dots$ be the indices where the parity changes in $a_0, a_1, a_2, \dots$, i.e., $i_0 = 0$ and $i_n > i_{n-1}$ is minimal such that $a_{i_n} - a_{i_{n-1}}$ is odd. Then the sum is
$$ \sum_{n=1}^\infty \frac{(-1)^{n-1}}{i_n}.$$
From the values given above, we have $i_1 = 1, i_2 = 2, i_3 \geq 5$, so the sum is in $[0.5, 0.7]$.
Unfortunately, computing the parity of $a_k$ is probably completely intractable, so it's very hard to say any more.
Things are much nicer if you use the binary log star, $\operatorname{lg}^*$, as then $a_0 = 1, a_1 = 2, a_2 = 2^2, \dots$ have obvious parity. We get $i_1 = 1$ and nothing further, so the sum is 1.
