# Infinite Series

## Key Questions

See below

#### Explanation:

There are different types of series, to what use different methods of evaluating

For example a converging geometric series:

$a + a r + a {r}^{2} + a {r}^{3} + \ldots + a {r}^{k} = {\sum}_{n = 1}^{k} a {r}^{n - 1}$

where ${\sum}_{n = 1}^{k} a {r}^{n - 1} = \frac{a \left(1 - {r}^{k}\right)}{1 - r}$

Assuming $| r | < 1$ we can let $k \to \infty$ for infinite series to be evaluated ...

${\lim}_{k \to \infty} {\sum}_{n = 1}^{k} a {r}^{n - 1} = {\lim}_{k \to \infty} \frac{a \left(1 - {r}^{k}\right)}{1 - k}$

as $k \to \infty$ , ${r}^{k} \to 0$ as $| r | < 1$

$\implies {\sum}_{n = 1}^{\infty} a {r}^{n - 1} = \frac{a}{1 - r}$

but there are other series what can be approached with tricks!

Take $\frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{30} + \ldots$

After consideration we can recognise this is the same as...

$\left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right) + \ldots$

$\left(\frac{1}{2} \cancel{- \frac{1}{3}}\right) + \left(\cancel{\frac{1}{3}} \cancel{- \frac{1}{4}}\right) + \left(\cancel{\frac{1}{4}} \cancel{- \frac{1}{5}}\right) + \ldots$

$= \frac{1}{2}$

There are also other infinite series that you can remember, and may be able to prove, a like:

e^x = 1 + x + x^2 /(2!) + x^3 / (3!) + ... = sum_(n=0) ^oo x^n / (n!)

There are many others, where there insist one set way of computing infinite series, there are many!

• I believe that it is the same as an alternating series. If that is the case, then an oscillating series is a series of the form:

${\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {b}_{n}$, where ${b}_{n} \ge 0$.

For example, the alternating harmonic series

${\sum}_{n = 1}^{\infty} \frac{{\left(- 1\right)}^{n}}{n}$

is a convergent alternating series.

• Here is an example of a telescoping series
${\sum}_{n = 1}^{\infty} \left(\frac{1}{n} - \frac{1}{n + 1}\right)$
$= \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots$
As you can see above, terms are shifted with some overlapping terms, which reminds us of a telescope. In order to find the sum, we will its partial sum ${S}_{n}$ first.
${S}_{n} = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n + 1}\right)$
by cancelling the overlapping terms,
$= 1 - \frac{1}{n + 1}$
Hence, the sume of the infinite series can be found by
${\sum}_{n = 1}^{\infty} \left(\frac{1}{n} - \frac{1}{n + 1}\right) = {\lim}_{n \to \infty} {S}_{n} = {\lim}_{n \to \infty} \left(1 - \frac{1}{n + 1}\right) = 1$

• It is very tough to answer such a general question, but I will give it a shot. Infinite series allow us to add up infinitely many terms, so it is suitable for representing something that keeps on going forever; for example, a geometric series can be used to find a fraction equivalent to any given repeating decimal such as:

$3.333 \ldots$

by splitting into individual decimals,

$= 3 + 0.3 + 0.03 + 0.003 + \cdots$

by rewriting into a form of geometric series,

$= 3 + 3 \left(\frac{1}{10}\right) + 3 {\left(\frac{1}{10}\right)}^{2} + 3 {\left(\frac{1}{10}\right)}^{3} + \cdots$

by using the formula for the sum of geometric series,

$= \frac{3}{1 - \frac{1}{10}} = \frac{10}{3}$

The knowledge of geometric series helped us find the fraction fairly easily. I hope that this was helpful.