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?
 A: 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 
$$  \frac{64}{3} \ln(2) - \frac{63}{4} \ln(3) +\frac{161}{36} \pi \sqrt{3}$$
A: 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
$$\Psi (p)=\log \left({p}\right)-\frac{1}{2 p}-\frac{1}{12 p^2}+\frac{1}{120
   p^4}+O\left(\frac{1}{p^6}\right)$$ making
$$S_m=\left(\frac{64}{3} \ln(2) - \frac{63}{4} \ln(3) +\frac{161}{36} \pi \sqrt{3}\right)-\frac{13}{2
   m}+\frac{23}{3 m^2}-\frac{311}{36 m^3}+\frac{499}{54 m^4}-\frac{36377}{3888
   m^5}+O\left(\frac{1}{m^6}\right)$$ Below are listed some values
$$\left(
\begin{array}{ccc}
 m & \text{approximation} & \text{exact} \\
 2 & 19.691108 & 19.787015 \\
 3 & 20.259944 & 20.269370 \\
 4 & 20.565282 & 20.567065 \\
 5 & 20.768485 & 20.768971 \\
 6 & 20.914700 & 20.914867 \\
 7 & 21.025135 & 21.025202 \\
 8 & 21.111527 & 21.111558 \\
 9 & 21.180965 & 21.180981
\end{array}
\right)$$
A: 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?
A: If you also want to calculate the sum of this series there's no need for anything but the series of the complex logarithm
$$
S(z)=-\log(1-z)=z+\frac{z^2}2+\frac{z^3}3+\cdots=\sum_{n=1}^\infty\frac{z^n}n
$$
evaluated at selected roots of unity $\neq1$. This is fine because the series $S(z)$
converges when $|z|\le1, z\neq1$.
Assume that the sequence $(a_n)_{n\ge1}$ is periodic with period $L$. Also assume that $a_1+a_2+\cdots+a_L=0$. We can then write
$$
\sum_{n=1}^\infty\frac{a_n}n
$$
as a linear combination of the series $S(\zeta_L^k)$, $k=1,2,\ldots,L_1$, with $\zeta_L=e^{2\pi i/L}$.
The tool for that is the discrete Fourier transform on $\Bbb{Z}_L.$ We have the characters 
$$\chi_j:\Bbb{Z}_L\to\Bbb{C}^*, \chi_j(\overline{n})=\zeta_L^{jn}, n=0,1,\ldots,L_1.$$
DFT guarantees that there exists coefficients $c_0,c_1,\ldots,c_{L-1}$ such that
for all $n=0,1,\ldots,L-1$
$$
a_n=\sum_{j=0}^{L-1}c_j\chi_j(\overline{n}).
$$
More precisely, the inverse Fourier transform gives the formula
$$
c_j=\frac1L\sum_{n=0}^{L-1}a_n\overline{\chi_j}(\overline{n})=\frac1L\sum_{n=0}^{L-1}a_n\zeta_L^{-nj}.
$$
Observe that 
$$c_0=\frac1L(a_1+a_2+\cdots+a_L)=0.$$
This will be important because it implies that the divergent harmonic series is missing in what follows, namely
$$
\sum_{n=1}^\infty\frac{a_nz^n}n=\frac1L\sum_{j=0}^{L-1}\sum_{n=1}^\infty \frac{c_j\zeta_L^{nj}z^n}n=\frac1L\sum_{j=0}^{L-1}c_jS(\zeta^{j}z)$$
whenever all the series converge. 
In the present case $L=6, a_1=1,a_2=10,a_3=100,a_4=a_5=a_6=-37$, so the zero assumption holds (as also pointed out by all the other answerers). A routine calculation gives
$$
\begin{aligned}
c_1=\overline{c_5}&=\frac1{12}(-283-85i\sqrt3)\\
c_2=\overline{c_4}&=\frac34(21+i\sqrt3)\\
c_3&=-\frac{64}3.
\end{aligned}
$$
So the sum of this series is
$$
-\frac16\sum_{j=1}^5c_j\log(1-\zeta_6^j).
$$
