# 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?

• Related: 1 and 2 – Mason Dec 13 '18 at 2:02

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?

• Alternatively, you can write $$F_n = {\frac {303264\,{n}^{4}+649296\,{n}^{3}+471096\,{n}^{2}+132516\,n+ 11892}{ \left( 6\,n+1 \right) \left( 6\,n+2 \right) \left( 6\,n+3 \right) \left( 6\,n+4 \right) \left( 6\,n+5 \right) \left( 6\,n+6 \right) }}$$ and note that the numerator has degree $4$ and denominator degree $6$, so $F_n = O(1/n^2)$. – Robert Israel Dec 12 '18 at 22:09

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 $$\frac{64}{3} \ln(2) - \frac{63}{4} \ln(3) +\frac{161}{36} \pi \sqrt{3}$$

• How is $\Psi$ defined exactly? – Karlo Dec 13 '18 at 10:13
• Also known as the Digamma function. $\Psi(x) = \dfrac{d}{dx} \ln(\Gamma(x))$. – Robert Israel Dec 13 '18 at 13:28

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

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 $$-\frac16\sum_{j=1}^5c_j\log(1-\zeta_6^j).$$

• A similar example. Too similar in fact. Turning this into CW. Likely that the same technique has been used in other answers as well. – Jyrki Lahtonen Dec 18 '18 at 19:56