Waste less time on Facebook — follow Brilliant.
×

Playing with Integrals: Limits and Integrals 2

I would like to continue the discussion started in this post by offering you a set of even more challenging limits. (As an additional exercise try solving these limits without Riemann Sums) Lets proceed to our first not so trivial example.

Problem 4. Find the following limit \[\lim_{n\to\infty}\frac{1}{n}\sqrt[n]{(n+1)(n+2)...(2n)}\]

Solution. We can't apply our tactic directly because we have to deal with product rather than sum. However, as many of you can guess, there is a natural way to transform the product into sum. \[\lim_{n\to\infty}\frac{1}{n}\sqrt[n]{(n+1)(n+2)...(2n)}=\lim_{n\to\infty}\sqrt[n]{\prod^n_{i=1}\left(1+\frac{i}{n}\right)}=\] \[=\lim_{n\to\infty}{\prod^n_{i=1}\left(1+\frac{i}{n}\right)}^{\frac{1}{n}}=\lim_{n\to\infty}e^{\frac{1}{n}\sum^n_{i=1}\ln\left(1+\frac{i}{n}\right)}\]

Now we see a familiar construction, which can be easily evaluated, namely \[\lim_{n\to\infty}\frac{1}{n}\sum^n_{i=1}\ln\left(1+\frac{i}{n}\right)=\int^1_0\ln(1+x)\,dx=\ln4 - 1.\]

Now returning to the initial problem we get \[\lim_{n\to\infty}\frac{1}{n}\sqrt[n]{(n+1)(n+2)...(2n)}=\lim_{n\to\infty}e^{\frac{1}{n}\sum^n_{i=1}\ln\left(1+\frac{i}{n}\right)}=\] \[=e^{\ln4 - 1}=\boxed{\frac{4}{e}}.\]

Almost the same problem was proposed at National Mathematics Olympiad in Moldova in 2007.

Problem 5. (Moldova, 2007, \(12^{th}\) class) Calculate the limit \[\lim_{n\to\infty}\frac{\sqrt[n+1]{(2n+3)(2n+4)...(3n+3)}}{n+1}\]

I also would like to propose another problem from Moldova, this time it's District Mathematics Olympiad, 2013. It's much easier than examples discussed in this post, but still serves a decent technique trainer.

Problem 6. (Moldova, 2013, \(12^{th}\) class) Find the limit \[\lim_{n\to\infty}\sum^n_{i=1}\frac{n}{(n+1)\sqrt{n(n+i)}}\]

Note by Nicolae Sapoval
3 years, 12 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

Nice post Nicolae! :)

Problem 5:

It can be written as:

\(\displaystyle \lim_{n\rightarrow \infty} \exp\left(\frac{1}{n+1}\sum_{i=1}^{n+1}\ln\left(2+\frac{i}{n+1}\right)\right)\)

The above is equivalent to:

\( \displaystyle \exp\left(\int_0^1 \ln(2+x)\,\,dx\right) = \exp\left(\ln\left(\frac{27}{4}\right)-1\right)=\frac{27}{4e}\)

Problem 6:

The given expression is equivalent to:

\(\displaystyle \lim_{n\rightarrow \infty} \sum_{i=1}^n \cfrac{1}{n\left(1+\frac{1}{n}\right)\sqrt{1+\frac{i}{n}}} \)

As \(n\rightarrow \infty\), \(1+1/n=1\) and the rest of the expression is same as:

\(\displaystyle \int_0^1 \frac{1}{\sqrt{1+x}}dx=2(\sqrt{2}-1)\)

Pranav Arora - 3 years, 11 months ago

Log in to reply

Good job! Stay tuned for the updates. I plan to write one more post like this and then to end the series on Limits and Integrals with a small test

Nicolae Sapoval - 3 years, 11 months ago

Log in to reply

Thank you. :)

I eagerly wait for your upcoming posts.

Pranav Arora - 3 years, 11 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...