How do you evaluate an infinite series?

Apr 8, 2018

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!