Convergence for a sequence #a_k# #(k=1,2,3,\ldots)# is the phenomenon that the terms #a_k# move closer and closer to a certain number #\alpha# as #k# becomes larger. The number #\alpha# is the limit to which the sequence converges. Here we will discuss convergence for geometric series.
A geometric sequence has the form #a_k=d\cdot c^{k-1}#, where #k=1,2,\ldots# and #c#, #d\in\mathbb{R}#.
Here, #c# is called the common ratio of the sequence.
The first term of the sequence is #d#. The first four terms are \[d,\ d\cdot c,\ d\cdot c^2,\ d\cdot c^3\]
If we only look at the first #n# terms for a natural number #n#, we call the sequence finite. To distinguish with the finite case, we also refer to the above geometric sequence as the infinite sequence.
The sequence #a_k = 6\cdot \frac{1}{3^k}# #(k=1,2,\ldots)# begins as follows.
\[2,\frac{2}{3},\ \frac{2}{9},\ \frac{2}{27},\ \frac{2}{81},\ \frac{2}{243},\ldots\]
This sequence is geometric because, for each #k#, the quotient #\dfrac{a_{k+1}}{a_k}# equals #\frac{1}{3}#, the common ratio of the sequence.
Indeed, the sequence can be written in the form of a geometric sequence:
\[a_k = 2\cdot \left(\frac{1}{3}\right)^{k-1}\]
The sequence #a_k = 3\cdot \frac{k^2}{k!}# #(k=1,2,\ldots)# begins as follows.
\[3,\ 6,\ \frac{9}{2},\ 2,\ \frac{5}{8},\ \frac{3}{20},\ \ldots\]
This sequence is not be geometric since the quotient #\dfrac{a_{k+1}}{a_k}# varies for different #k#.
A sequence that is built up by summing the initial terms #a_1,\ldots,a_n# for increasing #n# of a given sequence #a_k# #(k=1,2,\ldots)#, is called a series, or, more specifically, the series of the sequence #a_k# #(k=1,2,\ldots)#. Thus, the series #b_n# #(n=1,2,\ldots)# of #a_k# #(k=1,2,\ldots)# is given by \[ b_n=\sum_{k=1}^na_k\]
We are often interested in the existence and, if so, the value of the limit of the series #b_n# #(n=1,2,\ldots)# for #n\to\infty#. Therefore, the series is frequently presented as the infinite sum of the terms of the sequence #a_k# #(k=1,2,\ldots)#:
\[\lim_{n\to\infty}b_n =\lim_{n\to\infty} \sum_{k=1}^na_k=\sum_{k=1}^{\infty}a_k\]
Here we focus on the series of geometric sequences.
A geometric series is the series of a geometric sequence #a_k=d\cdot c^{k-1}#. Thus, the geometric series #b_n# #(n=1,2,\ldots)# pertaining to #a_k# #(k=1,2,\ldots)# is given by \[ b_n=\sum_{k=1}^na_k=\sum_{k=1}^nd\cdot c^{k-1}=d\sum_{k=1}^nc^{k-1}\]Again, #c# is called the common ratio of the series.
The series starts at #k=1# and the first four terms are \[d,\ d+d\cdot c,\ d+d\cdot c+d\cdot c^2,\ d+d\cdot c+d\cdot c^2+d\cdot c^3\]
The geometric sequence #a_k = 6\cdot \frac{1}{3^k}={ 2}\cdot \frac{1}{3^{k-1}}# #(k=1,2,\ldots)# begins as follows.
\[2,\frac{2}{3},\ \frac{2}{9},\ \frac{2}{27},\ \frac{2}{81},\ \frac{2}{243},\ldots\]
This corresponding series begins as follows.
\[{2},\ \frac{8}{3},\ \frac{26}{9},\ \frac{80}{27},\ \frac{242}{81},\ \frac{728}{243},\ldots\]
Instead of summing over #n# terms, we can calculate the #n#-th term of the geometric series directly by means of a simple formula.
For #c\ne1# we have \[\sum_{k=1}^nc^{k-1}=\sum_{k=0}^{n-1}c^k=\frac{c^n-1}{c-1}\]
We prove this formula by induction on #n#. We need to prove the formula\[\sum_{k=0}^{n-1}c^k=\dfrac{c^n-1}{c-1}\]If #n=1#, then the left-hand side is #c^0 = 1# and the right-hand side is #\dfrac{c^1-1}{c-1} = 1#. Because both sides are equal, the formula is correct for #n=1#.
Now let #n\ge1# and assume that the formula is correct for #n#. For #n+1# we have
\[\begin{array}{rcl}\displaystyle\sum_{k=0}^{n}c^k&=&\displaystyle\left(\sum_{k=0}^{n-1}c^k\right) + c^n\\ &&\phantom{xxx}\color{blue}{\text{last term separated from summation}}\\&=&\displaystyle \left(\dfrac{c^n-1}{c-1}\right)+ c^n\\&&\phantom{xxx}\color{blue}{\text{induction hypothesis}}\\&=&\displaystyle\dfrac{c^n-1+c^n(c-1)}{c-1}\\&&\phantom{xxx}\color{blue}{\text{expressions brought under a common denominator}}\\&=&\displaystyle \dfrac{c^n-1+c^{n+1}-c^n}{c-1}\\&&\phantom{xxx}\color{blue}{\text{numerator expanded}}\\&=&\displaystyle \dfrac{c^{n+1}-1}{c-1}\\&&\phantom{xxx}\color{blue}{\text{numerator simplified}}\\\end{array}\]
Thus, we derived the formula for #n+1#. In view of induction #n#, we conclude that the formula is correct for all #n\ge1#.
Using #3^6 = 729# and the finite geometric series formula, we can calculate the sum of the first six terms of the geometric sequence #6\cdot \frac{1}{3^k}# #(k=1,2,\ldots)# as follows.
\[2+\frac{2}{3}+ \frac{2}{9}+ \frac{2}{27}+ \frac{2}{81}+ \frac{2}{243}=6\sum_{k=1}^6 \frac{1}{3^{k}}=2\sum_{k=0}^5 \frac{1}{3^{k}} = 2\cdot\dfrac{\left(\frac{1}{3}\right)^6-1}{\frac{1}{3}-1}=\frac{728}{243} \]
The formula can be used to determine convergence of the infinite geometric series.
If #|c|\lt 1#, then the geometric series #1+c+c^2+\cdots # converges to \[\sum_{k=0}^\infty c^k=\dfrac{1}{1-c}\]
If #|c|\ge1#, then the series diverges.
The idea behind the proof is as follows. In case #|c|\lt 1#, we take the limit of the expression for the initial parts of the series given by
Finite geometric series formula in order to prove convergence and determine the limit. If the terms #c^n# do not tend to #0# when #n# goes to #\infty#, then the series must diverge. This case prevails if #|c|\ge 1#.
The Finite geometric series formula gives #\sum_{k=0}^{n-1}c^k = \dfrac{c^{n}-1}{c-1}# for #c\ne1#. Since #\lim_{n\to\infty}c^n = 0# for #|c|\lt 1#, we derive from this \[\begin{array}{rcl}\displaystyle \sum_{k=0}^\infty c^k&=&\displaystyle \lim_{n\to\infty}\sum_{k=0}^{n-1}c^k\\ &&\phantom{xxx}\color{blue}{\text{definition of infinite sum}}\\&=&\displaystyle \lim_{n\to\infty}\frac{c^n-1}{c-1}\\&&\phantom{xxx}\color{blue}{\text{finite geometric series formula}}\\&=&\displaystyle \frac{\left({\lim_{n\to\infty}c^n }\right)-1}{c-1}\\ && \phantom{xxx} \color{blue}{\text{continuity of the function }{(x-1)}/{(c-1)} \text{ of } x}\\ &=&\displaystyle \frac{0-1}{c-1}\\&&\phantom{xxx}\color{blue}{\lim_{n\to\infty}c^n = 0}\\&=&\displaystyle \frac{1}{1-c}\end{array}\]
Now suppose that #|c|\ge 1#. Assume that the series converges to a real number #\lambda#. Then there exists a natural number #N# such that, for all #n\gt N#, we have \[ \left|\lambda - \sum_{j=0}^n c^j\right| \lt \frac12\] Since #|c|\ge 1#, we have #|c|^n\ge1# for all #n#. Therefore, for each #k\gt N#,
\[\begin{array}{rcl}\dfrac12&\gt&\displaystyle \left|\lambda -\sum_{j=0}^{k+1} c^j\right|\\&&\phantom{xxx}\color{blue}{\text{above inequality applicable since }k+1\gt N}\\&\gt&\displaystyle \left|\left(\lambda -\sum_{j=0}^{k} c^j\right)-c^{k+1}\right|\\&&\phantom{xxx}\color{blue}{\text{term }c^{k+1}\text{ isolated}}\\&\ge&\displaystyle\left|c^{k+1} \right|-\left|\lambda -\sum_{j=0}^{k} c^j\right|\\&&\phantom{xxx}\color{blue}{\text{triangle inequality }\abs{a-b}\ge \abs{b}-\abs{a}}\\&\gt& 1-\frac12 \\&&\phantom{xxx}\color{blue}{|c|\ge 1\text{; above inequality applicable since }k\gt N}\\&=& \displaystyle\dfrac12\end{array}\]
We have found #\frac12\gt\frac12#. This contradiction shows that there is no real number #\lambda# to which the series converges. We conclude that the series diverges for #|c|\ge1#.
The infinite geometric series formula applied to the series of the geometric sequence #6\cdot \frac{1}{3^k}# #(k=1,2,\ldots)# gives
\[\sum_{k=1}^\infty6\cdot \frac{1}{3^k}=2\sum_{k=0}^\infty \frac{1}{3^{k}} = 2\cdot\dfrac{1}{1-\frac{1}{3}}= {3} \]
Above, we saw that the first six terms add up to #\frac{728}{243}#, which differs from the limit #{3}# by #\frac{1}{243}#. For general #n# the size of the difference between the limit and the sum of the first #n# terms of the sequence is equal to \[\left|\frac{1}{1-\frac{1}{3}}-\frac{1-\left(\frac{1}{3}\right)^n}{1-\frac{1}{3}}\right| = \frac{1}{3^{n-1}}\]
The graph of the function # \green{\frac{1}{1-x}}# of #x# is depicted below, as well as the finite geometric series #\blue{1+x+x^2+\cdots +x^n}# for different values of #n#.
Calculate the sum #\sum_{k=0}^{8}3^k#.
#\sum_{k=0}^{8}3^k=# #9841#
If #c\neq1#, then #\sum_{k=0}^{n-1}c^k=\frac{c^n-1}{c-1}#. Entering of #n=9# and #c=3# gives\[\sum_{k=0}^{8}3^k=\frac{3^{9}-1}{3-1}=9841\tiny.\]
Geometric series
