Xrfgtjtk

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
$$left(
begin{array}{ccc}
m & text{approximation} & text{exact} \
2 & 19.691108 & 19.787015 \
3 & 20.259944 & 20.269370 \
4 & 20.565282 & 20.567065 \
5 & 20.768485 & 20.768971 \
6 & 20.914700 & 20.914867 \
7 & 21.025135 & 21.025202 \
8 & 21.111527 & 21.111558 \
9 & 21.180965 & 21.180981
end{array}
right)$$

If you also want to calculate the sum of this series there's no need for anything but the series of the complex logarithm
$$
S(z)=-log(1-z)=z+frac{z^2}2+frac{z^3}3+cdots=sum_{n=1}^inftyfrac{z^n}n
$$
evaluated at selected roots of unity $neq1$. This is fine because the series $S(z)$
converges when $|z|le1, zneq1$.

Assume that the sequence $(a_n)_{nge1}$ is periodic with period $L$. Also assume that $a_1+a_2+cdots+a_L=0$. We can then write
$$
sum_{n=1}^inftyfrac{a_n}n
$$
as a linear combination of the series $S(zeta_L^k)$, $k=1,2,ldots,L_1$, with $zeta_L=e^{2pi i/L}$.

The tool for that is the discrete Fourier transform on $Bbb{Z}_L.$ We have the characters
$$chi_j:Bbb{Z}_LtoBbb{C}^*, chi_j(overline{n})=zeta_L^{jn}, n=0,1,ldots,L_1.$$
DFT guarantees that there exists coefficients $c_0,c_1,ldots,c_{L-1}$ such that
for all $n=0,1,ldots,L-1$
$$
a_n=sum_{j=0}^{L-1}c_jchi_j(overline{n}).
$$
More precisely, the inverse Fourier transform gives the formula
$$
c_j=frac1Lsum_{n=0}^{L-1}a_noverline{chi_j}(overline{n})=frac1Lsum_{n=0}^{L-1}a_nzeta_L^{-nj}.
$$
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
$$
sum_{n=1}^inftyfrac{a_nz^n}n=frac1Lsum_{j=0}^{L-1}sum_{n=1}^infty frac{c_jzeta_L^{nj}z^n}n=frac1Lsum_{j=0}^{L-1}c_jS(zeta^{j}z)$$
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
$$
begin{aligned}
c_1=overline{c_5}&=frac1{12}(-283-85isqrt3)\
c_2=overline{c_4}&=frac34(21+isqrt3)\
c_3&=-frac{64}3.
end{aligned}
$$
So the sum of this series is
$$
-frac16sum_{j=1}^5c_jlog(1-zeta_6^j).
$$