Series

Sum of arithmetic series

\[S_{n} = \frac{n * (a_{1} + a_{n})}{2}\]

Example

\[1 + 2 + 3 + ... + N\] \[S_{n} = \frac{N * (1 + N)}{2} = \frac{N * (N + 1)}{2}\]

Sum of geometric series

r < 1

Infinite sum of geometric series:

\[S_{n} = \frac{a_{0}}{1 - r}\]

Example

Calculate the sum of the series:

\[\sum_{n=0}^{∞} \frac{1}{2^{n}} = 1 + \frac{1}{2} + \frac{1}{4} + ...\]

We know that

\[a_{0} = \frac{1}{2^0} = 1\] \[r = \frac{1}{2}\]

So the infinite sum is:

\[S_{n} = \frac{\frac{1}{2^0}}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2\]

r > 1

\[S_{n} = a_{1} * \frac{r^n - 1}{r - 1}\]

Alternatively:

\[S_{n} = a_{1} * \frac{1 - r^n}{1 - r}\]

Example

Calculate the sum of the series

\[1 + 2 + 4 + 8 + ... + 1024\]

We know that \(1 + 2 + 4 + 8 + ... + 1024 = 2^{0} + 2^{1} + 2^{2} + 2^{3} + ... + 2^{10}\)

\[a_{1} = 1\] \[r = 2\] \[n = 11\]

(There are 11 elements in total because the series is: 2⁰ + 2¹ + 2² + 2³ + … + 2¹⁰)

So the sum is:

\[S_{n} = 1 * \frac{2^{11} - 1}{2 - 1} = \cancel{1} * \frac{2^{11} - 1}{\cancel{2 - 1}} = 2^{11} - 1\]

Sum of series

\[\sum_{i=1}^{n} c = c * n\] \[\sum_{i=1}^{n} i = \frac{n*(n+1)}{2}\] \[\sum_{i=1}^{n} i^{2} = \frac{n*(n+1)*(2n+1)}{6}\] \[\sum_{i=1}^{n} i^{3} = \frac{n^{2}*(n+1)^{2}}{4}\]