Expressing $\frac{13}{17}$ as an infinite series...

Question

I am trying to find a way to express $\frac{13}{17}$ as an infinite series so that I can convert $\frac{13}{17}$ to its base-$3$ counterpart. Could someone please how such an infinite series can be developed? How does one even start solving this problem?

Answer 1

Two options:

• Convert 13 and 17 to base 3 and then do long division in base 3, and then recognize when the process starts repeating
• Find an instance of $3^n - 1$ that is divisible by 17 to write your fraction as having a denominator of $3^n-1$. Note that $\frac1{3^n-1}$ has a simple representation in base 3 (just like $\frac1{10^n-1}$ has a simple representation in base 10).

Answer 2

Hint -

Step 1: Find the continued decimal expansion of 13/17.

Step 2: Convert this fraction to base-3, using the known rules for base conversion.

Answer 3

One way is to divide it out in base three:

        0.2021221102010011
     _____________________
  122)111.0000000000000000
      102 1
      -----
        1 200
        1 021
        -----
          1020
           122
          ----
           1210
           1021
           ----
            1120
            1021
            ----
              220
              122
              ---
               210
               122
               ---
                1100
                1021
                ----
                   200
                   122
                   ---
                     1000
                      122
                      ---
                      1010
                       122
                      ----
                       111

At this point we have the remainder with which we started, so the digits in the quotient will repeat, and the ternary expansion is

$$0.\overline{2021221102010011}\,.$$

Now

$$2021221102010011_{\text{three}}=32\,918\,080$$

and has $16$ digits in base three, so

$$\begin{align*} \frac{13}{17}&=\frac{32\,918\,080}{3^{16}}+\frac{32\,918\,080}{\left(3^{16}\right)^2}+\frac{32\,918\,080}{\left(3^{16}\right)^3}+\ldots\\\\ &=\sum_{n\ge 1}32\,918\,080\left(\frac1{3^{16}}\right)^n\,. \end{align*}$$

Answer 4

For finding an infinite series, we could try to assume that it has a geometric series representation. Recall that the sum of an infinite geometric series is:

$$S=\frac{u_1}{1-r}$$

where $u_1$ is the first term in the geometric series, and $r$ the common ratio between terms such that $|r|<1$. Suppose we take $u_1=\frac{1}{17}$. Then,

$$\frac{13}{17}=\frac{1}{17}\cdot\frac{1}{1-r}\\$$ $$13=\frac{1}{1-r}\\$$ $$1-r=\frac{1}{13}\\$$ $$\implies r=\frac{12}{13}\\$$

which is indeed less than 1 and positive. Therefore, we may write:

$$\frac{13}{17}=\frac{1}{17}\left(1+\frac{12}{13}+\left(\frac{12}{13}\right)^2+\dots\right)$$

or as a series,

$$\frac{13}{17}=\frac{1}{17}\sum_{n=0}^{\infty}\left(\frac{12}{13}\right)^n$$

Hope this helps!

Answer 5

It exists by standard results. For actually finding it, you can just use trial and error, comparing the remainder against multiples of powers of 1/3. $$\frac{13}{17} = \frac23 + \frac5{51} = \frac23 + \frac09 + \frac2{27} + \frac{11}{459} = \frac23 + \frac09 + \frac2{27} + \frac1{81} + \frac{16}{1377} = 0.2021\dots_3$$ Getting a computer to continue for me, we see the partial expansion

0.202122110201001120212211020100112021221102010011202122110201001120212211020100112021221102010011202

i.e.

$$\frac{13}{17} = 0.\overline{2021221102010011}_3$$

Basic python code:

from sympy import *
R=Rational(13,17)
ans='0.'
for i in range(1,100):
    if R-2*Rational(1,3**i)>=0:
        ans=ans+'2'
        R=R-2*Rational(1,3**i)
    elif R-Rational(1,3**i)>=0:
        ans=ans+'1'
        R=R-Rational(1,3**i)
    else:
        ans=ans+'0'
ans