Series

135. Series

\sum_{𝑖 = 1}^{\infty} 𝑎_{𝑖} = lim_{𝑛 \to \infty} \sum_{𝑖 = 1}^{𝑛} 𝑎_{𝑖}

Provided the limit exists:

If $𝑎_{𝑖} \geq 0$ : limit exists
If $𝑎_{𝑖}$ do not have the same sign:
- Limit need not exists
- Limit may exists but be different if we sum in a different order
  - Limit exists and independent of order of summation if

\sum_{𝑖 = 1}^{\infty} | 𝑎_{𝑖} | \leq \infty

Geometric Series

𝑆 = \sum_{𝑖 = 0}^{\infty} 𝛼^{𝑖} = 1 + 𝛼 + 𝛼^{2} + \dots = \frac{1}{1 - 𝛼} | 𝛼 | < 1

Derivation:

\begin{matrix} (1 - 𝛼) (1 + 𝛼 + \dots 𝛼^{𝑛}) = 1 - 𝛼^{𝑛 + 1} \\ 𝑛 \to \infty \\ (1 - 𝛼) 𝑆 = 1 \\ 𝑆 = \frac{1}{1 - 𝛼} \end{matrix}

Summation in series with multiple indices

\sum_{𝑖 \geq 1, 𝑗 \geq 1} 𝑎_{𝑖 𝑗}

If the sum of the absolute value of the terms of the series are finite,

\sum | 𝑎_{𝑖 𝑗} | < \infty

the order in which the elements are summed will not matter

E.g.:

\sum_{𝑖 = 1}^{\infty} (\sum_{𝑗 = 1}^{\infty} 𝑎_{𝑖 𝑗}) = \sum_{𝑗 = 1}^{\infty} (\sum_{𝑖 = 1}^{\infty} 𝑎_{𝑖 𝑗})