# Approximate value of a slowly-converging sum of $\sum|\sin n|^n/n$

In this question on Math.SE there appears this sum: $$S = \sum_{n\geq1}s_n, \qquad s_n = \frac{|\sin n|^n}{n},$$ which converges very slowly. What methods would you suggest for evaluating it numerically?

What I managed to do is unsatisfactory. Pick a bound $M$, and approximate the remainder of the sum over $n\geq M$ using $$\frac{|\sin n|^n}{n} \approx \frac1\pi\int_0^\pi \frac{|\sin t|^n}{n}\,dt = T_n,$$ then write $$S_M = \sum_{1\leq n<M}s_n + \sum_{n\geq M}T_n.$$ The infinite sum over $T_n$ has a closed form in terms of generalized hypergeometric functions.

I tried Shanks and Cohen-Villegas-Zagier convergence acceleration techniques (in mpmath) applied to $S_{10^6k}$ and $S_{10^7k}$, $1\leq k\leq10$, but they didn't work well, giving values about $10^{-3}$ away from each other ($2.151$). When I tried to fit $$S_M = S_\infty + b M^\alpha,$$ for $M=\{1,2,3\}\times 10^7$, I got $\alpha=-\frac12$ and $$S_\infty = 2.1509.$$

This is only five digits with $10^7$ terms. Is there a better method?

EDIT The values for $S_{10^6\{1\ldots10\}}$ are:

Z6 = [2.1503320981264467, 2.1504796881940735, 2.150563926321965,
2.150601535111673, 2.1506361337302553, 2.1506605927579456,
2.15067535745342, 2.150691134235195, 2.1507008825140925,
2.1507120293097395]


The values for $S_{10^7\{1\ldots10\}}$ are:

Z7 = [2.1507120293097395, 2.1507656995403974, 2.1507894767527613,
2.1508036508488018, 2.1508133237402696, 2.150820463977797,
2.1508260133696635, 2.1508304866069095, 2.150834191855564,
2.1508373250787898]


The values for $S_{1+10^8\{1\ldots10\}}$ are:

Z8 = [2.1508373250785855, 2.1508542694443973, 2.1508618019128343,
2.150866279387301, 2.150869344143939, 2.1508715999897547,
2.150873358367117, 2.1508747716998142, 2.1508759457833806,
2.1508769361805817]


The main reason I say that convergence acceleration seems to fail is that it gives somewhat different results (error $\sim 10^{-3}$) for $Z_6$ and $Z_7$. The simple model above gives the same result (to five digits), but introducing the next asymptotic term and trying to fit $S_\infty+b_1 M^{-1/2}+b_2 M^{-3/2}$ makes the two series agree slightly less, not more.

EDIT When fitting $S_\infty+b M^{-1/2}$ with least squares one thing to note is that the estimate for $Z_7$ lies two estimated standard errors from the estimate for $Z_6$, so I'm not sure how accurate it is. The better value, estimated from $Z_7$ seems to be $2.150895272(1)$, but this might be inaccurate.

EDIT The sum $\sum_{n\geq m}T_n$ behaves as $m^{-1/2}$ as $m\to\infty$, and is given by $$\frac{\Gamma(m)}{\pi 2^m}\left(\frac{2}{\Gamma(1+\frac m2)^2}F\left( \begin{array}{c} 1,\frac{1+m}{2},\frac m2\\1+\frac m2,1+\frac m2 \end{array}\right) + \frac{m}{\Gamma(\frac{3+m}{2})^2}F\left(\begin{array}{c} 1,\frac{1+m}{2},1+\frac m2\\ \frac{3+m}{2},\frac{3+m}{2}\end{array} \right) \right).$$

• Could you post the values of $S_M$ for $M=\{1,2,3,4,5,6,7,8,9,10\}\times 10^7$ ? Jun 22, 2014 at 4:25
• @Kirill Did they prove over on the other page that the sum was less than or equal to 2.15? Jun 22, 2014 at 7:41
• @ClaudeLeibovici I edited the question with the numbers. Jun 23, 2014 at 3:13
• @bobbym I'm not sure. Jun 23, 2014 at 4:24
• It seems that the next dominant error term in $S_M$ is coming from the tail of the sum over the $s_n$ rather than from the sum over the $T_n$. I wonder what kind of asymptotic series it has? Jun 23, 2014 at 5:07

For example to get the sum with an error not much above $$10^{-20}$$, I'd sum up only those values that make a contribution greater than $$10^{-20}/n$$. And that's quite simple, because $$|\sin(n)|^n$$ while be greater than $$10^{-20}$$ only if $$|\sin(n)|>10^{-20/n}$$ or if $$|\sin(n)|>1 - 46.05 /n$$, and for say $$n > 1,000,000$$, these $$n$$ are quite rare and could be found say using Euclid's algorithm - they are all integers $$n$$ close to odd multiples of $$\pi/2$$.
Addition: $$|\sin(n)|>1 - 46.05 /n$$ iff $$|\cos (n)|>9.6 n^{-1/2}$$, and that is about how close $$n$$ would have to be to an odd multiple of $$\pi/2$$. So this is less rare than I hoped - but still quite rare when $$n \geq 10^9$$, for example. And consecutive integers that are close to odd multiples of $$\pi/2$$ are very easy to find with a slight adaption of Euclid's algorithm. So the effort to calculate the sum up to n grows only with $$n^{1/2}$$, which means adding up to say $$n=10^{24}$$ or so is feasible.
PS. Doing the summation using long double (64 bit precision), there are only 2,296 values $$n$$ between $$30 \cdot 10^9$$ and $$30.1 \cdot 10^9$$ where $$\sin(n)$$ is close enough to $$\pm 1$$ so that the sum actually gets changed, that is less than one in 43,000 values. Picking only those values would make the calculation obviously a lot faster.