Waste less time on Facebook — follow Brilliant.
×

Limit.

Find the value of \( \displaystyle \lim_{n \to \infty} \left (\displaystyle \prod_{r = 1}^{n} \left ( 1 + \frac{r}{n} \right)\right)^{\frac{1}{n}}\)

Note by Kushagraa Aggarwal
4 years, 4 months ago

No vote yet
4 votes

  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

By properties of Riemann sums: \(\displaystyle\lim_{n \to \infty}\dfrac{1}{n}\sum_{r = 1}^{n}\ln \left(1 + \dfrac{r}{n}\right) = \int_{1}^{2}\ln(x)\,dx = \left[x\ln x - x\right]_{1}^{2} = 2\ln 2 - 1\).

Exponentiation yields \(\displaystyle\lim_{n \to \infty} \left[\prod_{r = 1}^{n}\left(1+\dfrac{r}{n}\right)\right]^{\tfrac{1}{n}} = e^{2 \ln 2 - 1} = \dfrac{4}{e}\).

Jimmy Kariznov - 4 years, 4 months ago

Log in to reply

Isn't the product kinda discrete. I mean r and n can only take integer values, while the limit as n approaches infinity kinda assumes n can take continuous values. Is there something wrrong with this belief?

Sebastian Garrido - 4 years, 4 months ago

Log in to reply

Hi,

He has mentioned there, Riemann sum

Gopinath No - 4 years, 4 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...