New user? Sign up

Existing user? Log in

Test the convergence

\[\large{\displaystyle \sum_{n=1}^{\infty} x^n}\]

if \(x>1\)

Note by Tanishq Varshney 3 years, 1 month ago

Easy Math Editor

*italics*

_italics_

**bold**

__bold__

- bulleted- list

1. numbered2. list

paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)

> 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"

2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

Sort by:

@Brian Charlesworth sir, @Satyajit Mohanty @Michael Mendrin sir

Log in to reply

Well, is it convergent?

dont know , was given homework

@Tanishq Varshney – Where do you study? Undergraduate or High School or Senior Secondary (+2) ?

@Satyajit Mohanty – under graduate, btw thanks for helping, are u sure it diverges

@Tanishq Varshney – Well Hmm, yup it's divergent! But anything can happen due to analytic continuations :D

For example \(\zeta(-2n) = 0\) \(\Rightarrow 1^{2n} + 2^{2n} + 3^{2n} + 4^{2n} + \ldots = 0\) for \(n \in \mathbb {Z}^+\)

I just know that:

\(\large{\displaystyle \sum_{n=1}^{\infty} ar^n}\) converges only when \(|r| < 1\), \(a \neq 0\). It diverges for \(|r| \geq 1\).

@Satyajit Mohanty – one more doubt is \(\displaystyle \sum_{r=1}^{\infty} r\) convergant or divergant. Becoz ramanujan proved this equal to -1/12

@Tanishq Varshney – \(\displaystyle \sum_{r=1}^\infty r \) is divergent. But again, Analytic Continuation changes the game and we find it as \(-\frac{1}{12}\).

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewest@Brian Charlesworth sir, @Satyajit Mohanty @Michael Mendrin sir

Log in to reply

Well, is it convergent?

Log in to reply

dont know , was given homework

Log in to reply

Log in to reply

Log in to reply

For example \(\zeta(-2n) = 0\) \(\Rightarrow 1^{2n} + 2^{2n} + 3^{2n} + 4^{2n} + \ldots = 0\) for \(n \in \mathbb {Z}^+\)

I just know that:

\(\large{\displaystyle \sum_{n=1}^{\infty} ar^n}\) converges only when \(|r| < 1\), \(a \neq 0\). It diverges for \(|r| \geq 1\).

Log in to reply

Log in to reply

Analytic Continuation changes the game and we find it as \(-\frac{1}{12}\).

\(\displaystyle \sum_{r=1}^\infty r \) is divergent. But again,Log in to reply