Argthtjtr

Consider the following series:

$$frac{1}{1} + frac{10}{2} + frac{100}{3} - frac{37}{4} - frac{37}{5} - frac{37}{6} + frac{1}{7} + frac{10}{8} + frac{100}{9} - frac{37}{10} - frac{37}{11} - frac{37}{12} + dots$$

This series seems to converge but I am not able to prove it. It seems impossible to write thing whole thing into summation notation. I tried to group by modulo 6, but then for example sum of $frac{1}{6n+1}$ diverges already so that doesn't seem right. Also tried to change the 37s into powers of 10 to maybe use the comparison test but also failed.
Any suggestions?

First, notice that $$F_n=frac{1}{6n+1}+frac{10}{6n+2}+frac{100}{6n+3}-frac{37}{6n+4}-frac{37}{6n+5}-frac{37}{6n+6} geq frac{111}{6n+3}-frac{111}{6n+4} > 0.$$
Then, notice that $$F_n leq frac{111}{6n+1}-frac{111}{6n+6}=frac{555}{(6n+1)(6n+6)}.$$

Thus the sum of $F_n$ converges. How can you infer the final result from this?

I presume the pattern $(1, 10, 100, -37, -37, -37)$ continues forever, so this is
$$ sum_{n=1}^infty frac{a_n}{n}$$
where $$ a_n = cases{1 & if $n equiv 1 mod 6$cr
10 & if $n equiv 2 mod 6$cr
100 & if $n equiv 3 mod 6$cr
-37 & if $n equiv 4,5$ or $0 mod 6$cr}$$
Note that $1+10+100 - 3times 37 = 0$. Then the $6m$
-th partial sum
$$ eqalign{S_m &= sum_{n=1}^{6m} frac{a_n}{n}cr
& = sum_{j=0}^m frac{1}{6j+1} +
10 sum_{j=0}^m frac{1}{6j+2} + 100 sum_{j=0}^m frac{1}{6j+3}
- 37 sum_{k=4}^6 sum_{j=0}^m frac{1}{6j+k}cr
&= frac{1}{6} left(Psi(m+1+1/6) - Psi(1/6) + 10 (Psi(m+1+2/6) - Psi(2/6)) + 100 (Psi(m+1+3/6) - Psi(3/6)) - 37 (Psi(m+1+4/6)+Psi(m+1+5/6)+Psi(m+1+6/6)-Psi(4/6)-Psi(5/6)-Psi(6/6))right)cr}$$

and since $Psi(x) = ln(x) + O(1/x)$ as $x to infty$, the sum converges.
In fact, the limit is
$$ frac{64}{3} ln(2) - frac{63}{4} ln(3) +frac{161}{36} pi sqrt{3}$$

Added for your curiosity but too long for a comment.

Starting from Robert Israel's elegant answer, it is possible to have quite accurate approximations of the partiel sums using the fact that, for large values of $p$ we have
$$Psi (p)=log left({p}right)-frac{1}{2 p}-frac{1}{12 p^2}+frac{1}{120
p^4}+Oleft(frac{1}{p^6}right)$$ making
$$S_m=left(frac{64}{3} ln(2) - frac{63}{4} ln(3) +frac{161}{36} pi sqrt{3}right)-frac{13}{2
m}+frac{23}{3 m^2}-frac{311}{36 m^3}+frac{499}{54 m^4}-frac{36377}{3888
m^5}+Oleft(frac{1}{m^6}right)$$ Below are listed some values
$$left(
begin{array}{ccc}
m & text{approximation} & text{exact} \
2 & 19.691108 & 19.787015 \
3 & 20.259944 & 20.269370 \
4 & 20.565282 & 20.567065 \
5 & 20.768485 & 20.768971 \
6 & 20.914700 & 20.914867 \
7 & 21.025135 & 21.025202 \
8 & 21.111527 & 21.111558 \
9 & 21.180965 & 21.180981
end{array}
right)$$