# series

$$S= \frac {1}{2\cdot4} + \frac {1\cdot3}{2\cdot4\cdot6} +\frac {1\cdot3\cdot5}{2\cdot4\cdot6\cdot8} \ldots n = ?$$

till n terms or till infinite

Note by Megh Parikh
4 years, 10 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Look at the binomial/Maclaurin expansion of $$(1-x)^{\frac12}$$ in the (valid) limit as $$x \to 1$$.

- 4 years, 10 months ago

till n terms $$[\frac{1}{2} - \frac{(2(n + 1)!)}{2^{2n +1}((n + 1)!)^2}]$$ , till $$\infty$$ $$\frac{1}{2}$$ ?

- 4 years, 10 months ago

Could you please give a solution?

- 4 years, 10 months ago

Observe that: $$\frac{1}{2.4} + \frac{1.3}{2.4.6} + \frac{1.3.5}{2.4.6.8} + ... n$$ terms

$$= \frac{\frac{1}{2}.\frac{1}{2}}{2!} + \frac{\frac{1}{2}.\frac{1}{2}.\frac{3}{2}}{3!} + ... n$$ terms

$$= [-1 - (\frac{1}{2})(-1) - \frac{(\frac{1}{2})(\frac{1}{2} - 1)}{2!}(-1)^2 - \frac{(\frac{1}{2})(\frac{1}{2}-1)(\frac{1}{2}-1)}{3!}(-1)^3 - ... n$$ terms $$] + \frac{1}{2}$$

Hence, this is $$(\frac{1}{2} - (1-x)^{\frac{1}{2}})$$ as $$x \rightarrow 1$$ with $$n \rightarrow \infty$$. At $$n=\infty$$, $$S=\frac{1}{2}$$

- 4 years, 10 months ago