A finite series is given by all the terms of a finite sequence, added together.
An infinite series is given by all the terms of an infinite sequence, added together.
To find the sum of an infinite series, start by summing the first few terms, i.e. $\sum^{2}_{k=1}$, $\sum^{3}_{k=1}$, $\sum^{4}_{k=1}$, $\dotso$. These sums of the first terms of the series are called partial sums. The $n^{\text{th} }$ partial sum of a series is the sum of the first $n$ terms.
If the partial sums tend towards a single number, this is called the limit and also the sum of the series.
What is the sum of the infinite series $\displaystyle{\sum^{\infty}_{k=1}\frac{1}{2^k
$? }}
Work out the first few partial sums:
\[\sum^{1}_{k=1}\frac{1}{2^k} = \frac{1}{2}\]
\[\sum^{2}_{k=1}\frac{1}{2^k} = \frac{1}{2}+\frac{1}{4}=\frac{3}{4}\]
\[\sum^{3}_{k=1}\frac{1}{2^k} = \frac{1}{2}+\frac{1}{4}+\frac{1}{8}=\frac{7}{8}\]
\[\sum^{4}_{k=1}\frac{1}{2^k} = \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=\frac{15}{16}\]
\[\sum^{5}_{k=1}\frac{1}{2^k} = \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}=\frac{31}{32}\]
Write down the partial sums as a sequence:
\[\frac{1}{2}, \frac{3}{4}, \frac{7}{8}, \frac{15}{16}, \frac{31}{32}, \dotso\]
We can see that the partial sums form a sequence that has a limit of $1$. So this series has sum $1$. Written as:
\[\sum^{\infty}_{k=1} \frac{1}{2^k} = 1\]
What is the sum of the infinite series $\displaystyle{\sum^{\infty}_{k=1}1}$?
Work out the first few partial sums:
\[\sum^{1}_{k=1}1 = 1\]
\[\sum^{2}_{k=1}1 = 1+1 =2\]
\[\sum^{3}_{k=1}1 = 1+1+1 =3\]
\[\sum^{4}_{k=1}1 = 1+1+1 +1 =4\]
Write down the partial sums as a sequence:
\[1, 2, 3, 4, \dotso\]
This sequence of partial sums does not tend to a real limit, it tends to infinity.
Therefore this series does not have a sum.