Is this Convergent?

Does this sum converge? If yes, then find its value.

Note by Danny Kills
3 years, 10 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. 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 1

paragraph 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

This is the "nested radical constant", also known as Vijayaraghavan's constant, which is \(1.75793275...\)

There is no exact expression for this constant.

Michael Mendrin - 3 years, 10 months ago

Log in to reply

Sir, how do we find out whether a sum like this converges or not? Thanks.

Satvik Golechha - 3 years, 10 months ago

Log in to reply

Look up "Herschfeld's Convergence Theorem". It states that for real terms \({ { x }_{ n } }\ge 0\) and real powers \(1>p>0\) if and only if

\(\displaystyle{ { x }_{ n } }^{ { p }^{ n } }\)

is bounded, then the following converges

\({ x }_{ 0 }+{ \left( { x }_{ 1 }+{ \left( { x }_{ 2 }+{ \left( ...+{ \left( { x }_{ n } \right) }^{ p } \right) }^{ p } \right) }^{ p } \right) }^{ p }\)

as \(n\rightarrow \infty \)

Here, \(p=\frac { 1 }{ 2 } \), so it's easy to see how the condition is met.

Michael Mendrin - 3 years, 10 months ago

Log in to reply

Comment deleted Aug 28, 2015

Log in to reply

@Naman Kapoor Is it a typo where \(r\) should be \(i\)? But even so, the terms approach \(1\) for large \(i\), so this doesn't converge.

Michael Mendrin - 3 years, 2 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...