Series convergence without sigma notation 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?

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 $1+10+100 - 3\times 37 = 0$.  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 
$$  \frac{64}{3} \ln(2) - \frac{63}{4} \ln(3) +\frac{161}{36} \pi \sqrt{3}$$
