Geometric Series

A term in geometric series is formed by multiplying the previous term by a constant - which is called the common ratio.

For example, the series 2, 4, 8, 16, 32, ... has a common ratio of 2.

Because, $\cfrac { 4 }{ 2 } =\cfrac { 8 }{ 4 } =\cfrac { 16 }{ 8 } =\cfrac { 32 }{ 16 } =2$.

In general, $r=\cfrac { { T }_{ 2 } }{ { T }_{ 1 } } =\cfrac { { T }_{ 3 } }{ { T }_{ 2 } } \\ \\ r=\cfrac { { T }_{ n } }{ { T }_{ n-1 } }$

Examples

1. Find x if 5 + x + 45 + ... is a geometric series.

For geometric series: $\cfrac { { T }_{ 2 } }{ { T }_{ 1 } } =\cfrac { { T }_{ 3 } }{ { T }_{ 2 } }$ $\cfrac { x }{ 5 } =\cfrac { 45 }{ x } \\ \\ { x }^{ 2 }=225\\ x=\pm \sqrt { 225 } \\ x=\pm 15$

If x = 15, the series is 5 + 15 + 45 + ...

If x = -15, the series is 5 - 15 + 45 - ...

2. Is $\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 6 } +\cfrac { 1 }{ 9 } +...$ a geometric series? $\cfrac { { T }_{ 2 } }{ { T }_{ 1 } } =\cfrac { 1 }{ 6 } \div \cfrac { 1 }{ 4 } \\ \\ \qquad =\cfrac { 1 }{ 6 } \times 4\\ \\ \qquad =\cfrac { 2 }{ 3 } \\ \\ \cfrac { { T }_{ 3 } }{ { T }_{ 2 } } =\cfrac { 1 }{ 9 } \div \cfrac { 1 }{ 6 } \\ \\ \qquad =\cfrac { 1 }{ 6 } \times 6\\ \\ \qquad =1\\ \\ \cfrac { { T }_{ 2 } }{ { T }_{ 1 } } \neq \cfrac { { T }_{ 3 } }{ { T }_{ 2 } }$

Therefore, the series is not geometric.

Terms of a Geometric Series Examples

1. Find the 9th term of the series 2 + 6 + 18 + ...

This is a geometric series with a = 2 and r = 3. ${ T }_{ n }=a{ r }^{ n-1 }\\ { T }_{ n }=2{ (3) }^{ n-1 }\\ For\quad the\quad 9th\quad term,\\ { T }_{ 9 }=2{ (3) }^{ 9-1 }\\ { T }_{ 9 }=1536$

2. Find the common ratio of $\cfrac { 2 }{ 3 } +\cfrac { 4 }{ 15 } +\cfrac { 8 }{ 75 } +...$ $r=\cfrac { 4 }{ 15 } \div \cfrac { 2 }{ 3 } (=\cfrac { 8 }{ 75 } \div \cfrac { 4 }{ 15 } )\\ \\ r=\cfrac { 4 }{ 15 } \times \cfrac { 3 }{ 2 } \\ \\ r=\cfrac { 2 }{ 5 }$

3. Which term of the series 4 + 12 + 36 + ... is equal to 78 732?

[altex]a=4,\quad r=3\\ \\ { T }_{ n }=a{ r }^{ n-1 }\\ \\ 78732=4{ (3) }^{ n-1 }\\ \\ 19683={ 3 }^{ n-1 }\\ \\ \log _{ 10 }{ 19683 } =\log _{ 10 }{ { 3 }^{ n-1 } } \\ \\ \log _{ 10 }{ 19683 } =(n-1)\log _{ 10 }{ 3 } \\ \\ \cfrac { \log _{ 10 }{ 19683 } }{ \log _{ 10 }{ 3 } } =n-1\\ \\ 9=n-1\\ \\ n=10[/latex]

So the 10th term is 78 732.

4. The second term of a geometric series is 6 and the 5th term is 162. Find the first term and common ratio. $a=?,\quad r=?\\ \\ { T }_{ n }=a{ r }^{ n-1 }\\ \\ { T }_{ 2 }=a{ r }^{ 2-1 }\\ a{ r }=6\qquad \qquad (1)\\ \\ { T }_{ 5 }=a{ r }^{ 5-1 }\\ a{ r }^{ 4 }=162\qquad \qquad (2)\\ \\ (2)\div (1):\\ { r }^{ 3 }=27\\ r=3\\ \\ a{ \left( 3 \right) }=6\\ a=2$

Partial Sum of a Geometric Series

The sum of the first n terms of a geometric series (nth partial sum) is given by the formula: ${ S }_{ n }=\cfrac { a\left( { r }^{ n }-1 \right) }{ r-1 } \quad for\quad \left| r \right| >1\\ \\ { S }_{ n }=\cfrac { a\left( 1-{ r }^{ n } \right) }{ 1-r } \quad for\quad \left| r \right| <1$

Examples

1. Find the sum of the first 10 terms of the series 3 + 12 + 48 ... $a=3,\quad r=4,\quad n=10\\ \\ { S }_{ 10 }=\cfrac { 3\left( { 4 }^{ 10 }-1 \right) }{ 4-1 } \\ \\ { S }_{ 10 }=\cfrac { 3\left( { 4 }^{ 10 }-1 \right) }{ 3 } \\ \\ { S }_{ 10 }={ 4 }^{ 10 }-1\\ { S }_{ 10 }=1048575$

2. Evaluate $\sum _{ 3 }^{ 11 }{ { 2 }^{ n } }$. $\sum _{ 3 }^{ 11 }{ { 2 }^{ n } } ={ 2 }^{ 3 }+{ 2 }^{ 4 }+...+{ 2 }^{ 11 }\\ \\ \qquad \quad \quad =8+16+...+2048\\ \\ a=8,\quad r=2,\quad n=9\\ \\ { S }_{ n }=\cfrac { 8\left( 2^{ 9 }-1 \right) }{ 2-1 } \\ \\ { S }_{ n }=8(512)\\ \\ { S }_{ n }=4088\\ \\$

3. Evaluate $40+10+2\cfrac { 1 }{ 2 } +...+\cfrac { 5 }{ 512 }$. $40+10+2\cfrac { 1 }{ 2 } +...+\cfrac { 5 }{ 512 } \\ \\ a=40,\quad r=\cfrac { 1 }{ 4 } ,\quad { T }_{ n }=\cfrac { 5 }{ 512 } \\ \\ Find\quad n:\\ \\ { T }_{ n }=a{ r }^{ n-1 }=\cfrac { 5 }{ 512 } \\ \\ 40{ \left( \cfrac { 1 }{ 4 } \right) }^{ n-1 }=\cfrac { 5 }{ 512 } \\ \\ { \left( \cfrac { 1 }{ 4 } \right) }^{ n-1 }=\cfrac { 5 }{ 20480 } \\ \\ \cfrac { 1 }{ { 4 }^{ n-1 } } =\cfrac { 1 }{ 4096 } \\ \\ \cfrac { 1 }{ { 4 }^{ n-1 } } =\cfrac { 1 }{ { 4 }^{ 6 } } \\ \\ n-1=6\\ \\ n=7$ ${ S }_{ n }=\cfrac { a\left( 1-{ r }^{ n } \right) }{ 1-r } \\ \\ { S }_{ 7 }=\cfrac { 40\left( 1-{ \left( \cfrac { 1 }{ 4 } \right) }^{ 7 } \right) }{ 1-\cfrac { 1 }{ 4 } } \\ \\ { S }_{ 7 }=\cfrac { 40\left( 1-\cfrac { 1 }{ 16384 } \right) }{ \cfrac { 3 }{ 4 } } \\ \\ { S }_{ 7 }=40\left( \cfrac { 16383 }{ 16384 } \right) \times \cfrac { 4 }{ 3 } \\ \\ { S }_{ 7 }=53\cfrac { 169 }{ 512 }$

Limiting Sum

In some geometric series, the sum becomes very large as n increases. This series diverges, or it has an infinite sum. However, some geometric  series has a limiting sum, that is, there is a limit to the sum of all of the terms in the series up to infinity-th term. Yes, infinity! This series converges and has a limiting sum.

For example, the series $2+1+\cfrac { 1 }{ 2 } +\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 8 } +...$, notice that after a while the terms are becoming closer and closer to zero and so will not add much to the sum of the whole series. Estimate its limiting sum. ${ S }_{ n }=\cfrac { a\left( 1-{ r }^{ n } \right) }{ 1-r }$

For $\left| r \right| <1$, as n increases, ${ r }^{ n }$ decreases and approaches zero e.g. when $r=\cfrac { 1 }{ 2 }$. $\therefore \quad { S }_{ \infty }=\cfrac { a(1-0) }{ 1-r } \\ \\ \qquad { S }_{ \infty }=\cfrac { a }{ 1-r }$

Using the example above, $a=2,\quad r=\cfrac { 1 }{ 2 } \\ \\ { S }_{ \infty }=\cfrac { 2 }{ 1-\cfrac { 1 }{ 2 } } \\ \\ { S }_{ \infty }=\cfrac { 2 }{ \cfrac { 1 }{ 2 } } \\ \\ { S }_{ \infty }=4$

Example

Evaluate $\sum _{ 2 }^{ \infty }{ 2 } { \left( \cfrac { 2 }{ 3 } \right) }^{ n }$ $\sum _{ 2 }^{ \infty }{ 2 } { \left( \cfrac { 2 }{ 3 } \right) }^{ n }=2{ \left( \cfrac { 2 }{ 3 } \right) }^{ 2 }+2{ \left( \cfrac { 2 }{ 3 } \right) }^{ 3 }+2{ \left( \cfrac { 2 }{ 3 } \right) }^{ 4 }+...\\ \\ \qquad \qquad \qquad =2\left( \cfrac { 4 }{ 9 } \right) +2\left( \cfrac { 8 }{ 27 } \right) +2\left( \cfrac { 16 }{ 81 } \right) +...\\ \\ \qquad \qquad \qquad =\cfrac { 8 }{ 9 } +\cfrac { 16 }{ 27 } +\cfrac { 32 }{ 81 } +...\\ \\ \\ a=\cfrac { 8 }{ 9 } ,\quad r=\cfrac { 2 }{ 3 } \\ \\ { S }_{ \infty }=\cfrac { a }{ 1-r } \\ \\ { S }_{ \infty }=\cfrac { \cfrac { 8 }{ 9 } }{ 1-\cfrac { 2 }{ 3 } } \\ \\ { S }_{ \infty }=\cfrac { 8 }{ 9 } \times \cfrac { 3 }{ 1 } \\ \\ { S }_{ \infty }=\cfrac { 24 }{ 9 } =2\cfrac { 2 }{ 3 }$