×

# How to start?

$$\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^2} \cos\Big(\dfrac{9}{n\pi + \sqrt{(n\pi)^2 - 9}}\Big) = - \dfrac{\pi^2}{12e^3}$$

Note by U Z
3 years, 2 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} $$\boxed{123}$$

## Comments

Sort by:

Top Newest

Can you approximate $$n \pi - \sqrt{ ( n \pi ) ^2 - 9 }$$ as $$n$$ gets large? Use Big-O notation if you are familiar with that.

Are you sure that there is an $$e^3$$ in there?

Also, please use distinct values for your indices. You cannot take the sum of the variable $$n$$ , as it goes from 1 to $$n$$.

Staff - 3 years, 2 months ago

Log in to reply

I am not familiar with that , I am sure that this is the question . Unable to solve that's why shared it and mentioned some brilliant members.

In a magazine - Mathematics today I saw it , Chapter - definite integration , section - Sandwich theorem

- 3 years, 2 months ago

Log in to reply

Shivang Jindal Ronak Agarwal Pratik Shastri Sudeep Salgia @Rube

- 3 years, 2 months ago

Log in to reply

What is p?

- 3 years, 2 months ago

Log in to reply

sorry its $$\pi$$

- 3 years, 2 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...