# Infinite Series

What is a Series
12:24 — by patrickJMT

Tip: This isn't the place to ask a question because the teacher can't reply.

## Key Questions

• It is very tough to answer such a general question, but I will give it a shot. Infinite series allow us to add up infinitely many terms, so it is suitable for representing something that keeps on going forever; for example, a geometric series can be used to find a fraction equivalent to any given repeating decimal such as:

$3.333 \ldots$

by splitting into individual decimals,

$= 3 + 0.3 + 0.03 + 0.003 + \cdots$

by rewriting into a form of geometric series,

$= 3 + 3 \left(\frac{1}{10}\right) + 3 {\left(\frac{1}{10}\right)}^{2} + 3 {\left(\frac{1}{10}\right)}^{3} + \cdots$

by using the formula for the sum of geometric series,

$= \frac{3}{1 - \frac{1}{10}} = \frac{10}{3}$

The knowledge of geometric series helped us find the fraction fairly easily. I hope that this was helpful.

The key step is to determine if the term a_n in the infinite series is one side bounded or not (in the limit).

Assume that the infinite series is given by ${\left\{{a}_{n}\right\}}_{n = 1}^{\infty}$, i.e., ${a}_{n}$ is the n-th term in the series.
1. First step is to check if the term is bounded by a positive finite value for all n
2. Second, check if the term reaches a finite constant in the limit i.e.,
${\lim}_{n \to \infty} {a}_{n} = c$
More precisely, for a given $\epsilon > 0$ (a very small positive real number) there exists a integer N such that
$| {a}_{n} - c | < \epsilon$ if $n > N$.
This basically means ${a}_{n}$ approaches c as n increases.

If a series satisfies this test, then it is convergent.
For example ${a}_{n} = \frac{1 + \frac{2}{n}}{2 + \frac{3}{n} ^ 2}$ is convergent because, for all n, terms a_n are smaller than or equal to $c = \frac{1}{2}$.
If this test fails, ie., there is no finite term c by which ${a}_{n}$ is bounded then the series is divergent.
For instance, ${a}_{n} = {e}^{n} / {n}^{2}$ is divergent as its values increases without bound.
Finally if the series is said to oscillate if the value fluctuates between two extremes.
eg. ${a}_{n} = {\left(- 1\right)}^{n}$.
The values are -1 and 1 for odd and even n's respectively.

Reference:
Chapter-section 5.1:
A. Khuri, Advanced calculus with applications in statistics, 2nd ed, Wiley, 2003, pp. 132 - 134.

• I believe that it is the same as an alternating series. If that is the case, then an oscillating series is a series of the form:

${\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {b}_{n}$, where ${b}_{n} \ge 0$.

For example, the alternating harmonic series

${\sum}_{n = 1}^{\infty} \frac{{\left(- 1\right)}^{n}}{n}$

is a convergent alternating series.

• Here is an example of a telescoping series
${\sum}_{n = 1}^{\infty} \left(\frac{1}{n} - \frac{1}{n + 1}\right)$
$= \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots$
As you can see above, terms are shifted with some overlapping terms, which reminds us of a telescope. In order to find the sum, we will its partial sum ${S}_{n}$ first.
${S}_{n} = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n + 1}\right)$
by cancelling the overlapping terms,
$= 1 - \frac{1}{n + 1}$
Hence, the sume of the infinite series can be found by
${\sum}_{n = 1}^{\infty} \left(\frac{1}{n} - \frac{1}{n + 1}\right) = {\lim}_{n \to \infty} {S}_{n} = {\lim}_{n \to \infty} \left(1 - \frac{1}{n + 1}\right) = 1$

## Questions

• · 1 week ago
• · 5 months ago
• · 6 months ago
• · 8 months ago
• · 10 months ago
• · 10 months ago
• · 11 months ago
• · 1 year ago
• · 1 year ago
• · 2 years ago
• · 2 years ago
• · 3 years ago
• · 3 years ago
• · 3 years ago
• · 3 years ago
• · 3 years ago
• · 3 years ago
• · 3 years ago
• · 3 years ago
• · 3 years ago
• · 3 years ago
• · 3 years ago
• · 3 years ago