Infinite Series

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Geometric Series

A geometric series is a series where the ratio between successive terms is constant. You can view a geometric series as a series with terms that form a geometric sequence (see the previous module on sequences).

For example, the series $$ \sum_{i=0}^\infty\left(\frac{1}{3}\right)^i=1+\frac13+\frac19+\frac{1}{27}+\ldots $$ is geometric with ratio $r=\frac13$.

Geometric series are our favorite series. It is always possible, and even easy, to determine whether a geometric series converges, and if it does, its value. This is rarely possible with other types of series.

The geometric series $\displaystyle\sum_{i=0}^\infty a r^i=a+ar+ar^2+ar^3 + \ldots$ converges to $\displaystyle\frac{a}{1-r}$ if $-1<r<1$ and diverges otherwise. Warning: this value of the series is true only when the series begins with $i=0$, so that the first term is $a$.

We could also say a geometric series $\sum ar^i$ converges if $|r|<1$ which is the same as $-1<r<1$. The absolute value inequality $|r|>1$ is equivalent to $r<-1$ or $r>1$.

This video will use the $n^{th}$ partial sums to show why we know the covergence and the value of geometric series.

The main issue is to confirm that a series is indeed geometric, and if so, what the values of $a$ and $r$ are.

Example: $5-\tfrac {10}{3}+\tfrac{20} {9}-\tfrac{40}{27}+\tfrac{80}{81}+\cdots.$

Solution: First, we try to determine if it is a geometric series. Dividing each term by the previous term shows that we have a common ratio of $r=-\tfrac 2 3$, with $a=a\cdot r^0=5$. So we have a geometric series, and since $|r|<1$, the series converges. Its value is $\frac{a}{1-r}=\frac{5}{1+\tfrac 2 3}=\frac{5}{\tfrac 5 3}=3$.
We have determined that $\displaystyle\sum_{n=0}^\infty 5\left(-\frac{2}{3}\right)^n=3$.

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Example: $3-\tfrac 6 5 + \tfrac{12}{25}-\tfrac{24}{125}+\cdots$.

DO: Find $r$ and $a$, and evaluate this series.

Solution: $\displaystyle\sum_{n=0}^\infty3\cdot\left(-\frac{2}{5}\right)^n=\frac{15}{7}$.

Integration by Parts

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Geometric Series