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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?

Markus Punnar
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4 Answers4

15

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?

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

Robert Israel
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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 $$\Psi (p)=\log \left({p}\right)-\frac{1}{2 p}-\frac{1}{12 p^2}+\frac{1}{120 p^4}+O\left(\frac{1}{p^6}\right)$$ 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 $$\left( \begin{array}{ccc} m & \text{approximation} & \text{exact} \\ 2 & 19.691108 & 19.787015 \\ 3 & 20.259944 & 20.269370 \\ 4 & 20.565282 & 20.567065 \\ 5 & 20.768485 & 20.768971 \\ 6 & 20.914700 & 20.914867 \\ 7 & 21.025135 & 21.025202 \\ 8 & 21.111527 & 21.111558 \\ 9 & 21.180965 & 21.180981 \end{array} \right)$$

Claude Leibovici
  • 260,315
0

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

Jyrki Lahtonen
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