Series

Topics: Calculus - Sequence

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(definition)

Let $\{a_n{\}}_{n\in \mathbb{N}}\subseteq \mathbb{R}$ be a sequence.

We define the partial sum sequence of $\{a_n\}$ as:

$$\left\{S_n\right\}={\left\{\sum _{i=1}^n{a}_i\right\}}_{n\in \mathbb{N}}$$

If this sequence converges, then we say that the series ${\sum }_{n=1}^{\infty }{a}_n$ converges and that ${\sum }_{n=1}^{\infty }{a}_n={\lim }_{n\to \infty }{S}_n$.

Example

Let $\{a_n\}=\left\{\frac{1}{2^{n-1}}\right\}$. It’s clear that:

• $a_1=1$
• $a_2=\frac{1}{2}$
• $a_3=\frac{1}{2^2}$
• $a_4=\frac{1}{2^3}$
• etc.

Its partial sum sequence $\{S_n\}$ gives us:

• $S_1=1$
• $S_2=1+\frac{1}{2}$
• $S_3=1+\frac{1}{2}+\frac{1}{2^2}$
• $S_4=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}$
• $⋮$
• $S_n=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots +\frac{1}{2^{n-1}}$

…so it becomes clear that:

$$\{S_n\}=\left\{\sum _{i=1}^n\frac{1}{2^{n-1}}\right\}$$

…which we can rewrite as:

$$\{S_n\}=\left\{\frac{1-\frac{1}{2^n}}{1-\frac{1}{2}}\right\}=\left\{2-\frac{1}{2^{n-1}}\right\}$$

Then, by calculating the limit of the partial sum sequence:

$$\underset{n\to \infty }{\lim }{S}_n=2$$

…we can obtain the concrete value to which the series converges:

$$\sum _{n=1}^{\infty }\frac{1}{2^{n-1}}=2$$

The convergence of series can be tested with several methods.