Geometric Series
Solved Problems
Example 3.
Find the sum of the series
\[{S_7} = 1 - \frac{1}{{\sqrt 2 }} + \frac{1}{2} - \frac{1}{{2\sqrt 2 }} + \frac{1}{4} - \frac{1}{{4\sqrt 2 }} + \frac{1}{8}.\]
Solution.
This is a geometric progression with \(q = - {\frac{1}{{\sqrt 2 }}}.\) Since the sum of a geometric progression is given by
\[{S_n} = {a_1}\frac{{1 - {q^n}}}{{1 - q}},\]
we have
\[{S_7} = 1 - \frac{1}{{\sqrt 2 }} + \frac{1}{2} - \frac{1}{{2\sqrt 2 }} + \frac{1}{4} - \frac{1}{{4\sqrt 2 }} + \frac{1}{8} = \frac{{1 - {{\left( { - \frac{1}{{\sqrt 2 }}} \right)}^7}}}{{1 - \left( { - \frac{1}{{\sqrt 2 }}} \right)}} = \frac{{1 - \frac{1}{{8\sqrt 2 }}}}{{1 + \frac{1}{{\sqrt 2 }}}} = \frac{{\frac{{8\sqrt 2 - 1}}{{8\sqrt 2 }}}}{{\frac{{\sqrt 2 + 1}}{{\sqrt 2 }}}} = \frac{{8\sqrt 2 - 1}}{{8\left( {\sqrt 2 + 1} \right)}}.\]
Example 4.
Express the repeating decimal \(0,131313 \ldots \) as a rational number.
Solution.
We can write:
\[0,131313 \ldots = \frac{{13}}{{100}} + \frac{{13}}{{10000}} + \frac{{13}}{{1000000}} + \ldots = \frac{{13}}{{100}}\left( {1 + \frac{1}{{100}} + \frac{1}{{10000}} + \ldots } \right).\]
Using the formula for the sum of infinite geometric series
\[S = \sum\limits_{n = 0}^\infty {{q^n}} = {\frac{1}{{1 - q}}}\]
with ratio \(q = \frac{1}{{100}},\) we obtain
\[0,131313 \ldots = \frac{{13}}{{100}} \cdot \frac{1}{{1 - \frac{1}{{100}}}} = \frac{{13}}{{100}} \cdot \frac{1}{{\frac{{99}}{{100}}}} = \frac{{13}}{{99}}.\]
Example 5.
Show that
\[1 + \frac{1}{x} + \frac{1}{{{x^2}}} + \frac{1}{{{x^3}}} + \frac{1}{{{x^4}}} + \ldots = \frac{x}{{x - 1}}\]
assuming \(x \gt 1.\)
Solution.
Note that if \(x \gt 1,\) then \({\frac{1}{x}} \lt 1.\) In this case, the left side is the sum of an infinite geometric progression. Using the formula
\[S = \sum\limits_{n = 0}^\infty {{q^n}} = {\frac{1}{{1 - q}}},\]
we can write the left side as
\[1 + \frac{1}{x} + \frac{1}{{{x^2}}} + \frac{1}{{{x^3}}} + \frac{1}{{{x^4}}} + \ldots = \frac{1}{{1 - \frac{1}{x}}} = \frac{1}{{\frac{{x - 1}}{x}}} = \frac{x}{{x - 1}},\]
so that the formula is proved.
Example 6.
Solve the equation
\[{x^2} - 2{x^3} + 4{x^4} - 8{x^5} + \ldots = 2x + 1,\;\;\left| x \right| \lt 1.\]
Solution.
We can write the left side of the equation using the formula for the sum of an infinite geometric series:
\[S = \sum\limits_{n = 0}^\infty {{q^n}} = \frac{1}{{1 - q}},\]
\[{x^2} - 2{x^3} + 4{x^4} - 8{x^5} + \ldots = {x^2}\left( {1 - 2x + 4{x^2} - 8{x^3} + \ldots } \right) = {x^2} \cdot \frac{1}{{1 - \left( { - 2x} \right)}} = \frac{{{x^2}}}{{1 + 2x}}.\]
Then
\[\frac{{{x^2}}}{{1 + 2x}} = 2x + 1,\;\; \Rightarrow {x^2} = {\left( {2x + 1} \right)^2},\;\; \Rightarrow {x^2} = 4{x^2} + 4x + 1,\;\; \Rightarrow 3{x^2} + 4x + 1 = 0.\]
The roots of the quadratic equations are
\[D = {4^2} - 4 \cdot 3 = 4,\;\; \Rightarrow {x_{1,2}} = \frac{{ - 4 \pm \sqrt 4 }}{6} = \frac{{ - 4 \pm 2}}{6} = - 1, - \frac{1}{3}.\]
Since \(\left| x \right| \lt 1,\) the answer is \(x = - {\frac{1}{3}}.\)
Example 7.
The second term of an infinite geometric progression (\(\left| q \right| \lt 1\)) is \(21\) and the sum of the progression is \(112.\) Determine the first term and ratio of the progression.
Solution.
We use the formula for the sum of an infinite geometric series:
\[S = \sum\limits_{n = 0}^\infty {{a_1}{q^n}} = \frac{{{a_1}}}{{1 - q}}.\]
Since the second term of a geometric progression is equal to \({a_2} = {a_1}q,\) we have the following system of equations to find the first term \({a_1}\) and ratio \(q:\)
\[
\left\{ \begin{array}{l}
112 = \frac{{{a_1}}}{{1 - q}}\\
21 = {a_1}q
\end{array} \right.\;\;\text{or}\;\;
\left\{ \begin{array}{l}
112 = \frac{{{a_1}}}{{1 - q}}\\
{a_1} = \frac{{21}}{q}
\end{array} \right..\]
Solving this system we obtain the following quadratic equation
\[112 = \frac{{\frac{{21}}{q}}}{{1 - q}},\;\; \Rightarrow 112 = \frac{{21}}{{q\left( {1 - q} \right)}},\;\; \Rightarrow 21 = 112q\left( {1 - q} \right),\;\; \Rightarrow 21 = 112q - 112{q^2},\;\; \Rightarrow 112{q^2} - 112q + 21 = 0,\;\; \Rightarrow 16{q^2} - 16q + 3 = 0.\]
The equation has two roots:
\[D = {\left( { - 16} \right)^2} - 4 \cdot 16 \cdot 3 = 256 - 192 = 64,\;\; \Rightarrow {q_{1,2}} = \frac{{16 \pm \sqrt {64} }}{{32}} = \frac{{16 \pm 8}}{{32}},\;\; \Rightarrow {q_1} = \frac{{24}}{{32}} = \frac{3}{4},\;\; {q_2} = \frac{8}{{32}} = \frac{1}{4}.\]
For each ratio \(q\) we determine the first terms:
\[\left( {{a_1}} \right)_1 = \frac{{21}}{{{q_1}}} = \frac{{21}}{{\frac{3}{4}}} = 28,\;\; \left( {{a_1}} \right)_2 = \frac{{21}}{{{q_2}}} = \frac{{21}}{{\frac{1}{4}}} = 84.\]
Thus, the problem has two answers:
- \({a_1} = 28,\;q = {\frac{3}{4}} ;\)
- \({a_1} = 84,\;q = {\frac{1}{4}} .\)
