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→∞1n(n+1)(n+2)...(2n)n\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→∞1n(n+1)(n+2)...(2n)n=lim⁡n→∞∏i=1n(1+in)n=\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→∞∏i=1n(1+in)1n=lim⁡n→∞e1n∑i=1nln⁡(1+in)=\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→∞1n∑i=1nln⁡(1+in)=∫01ln⁡(1+x) dx=ln⁡4−1.\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→∞1n(n+1)(n+2)...(2n)n=lim⁡n→∞e1n∑i=1nln⁡(1+in)=\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)}= =eln⁡4−1=4e.=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, 12th12^{th} class) Calculate the limit lim⁡n→∞(2n+3)(2n+4)...(3n+3)n+1n+1\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, 12th12^{th} class) Find the limit lim⁡n→∞∑i=1nn(n+1)n(n+i)\lim_{n\to\infty}\sum^n_{i=1}\frac{n}{(n+1)\sqrt{n(n+i)}}

#Calculus #Limits #Integration #Goldbach'sConjurersGroup

Note by Nicolae Sapoval
7 years, 5 months ago

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Comments

Nice post Nicolae! :)

Problem 5:

It can be written as:

lim⁡n→∞exp⁡(1n+1∑i=1n+1ln⁡(2+in+1))\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:

exp⁡(∫01ln⁡(2+x)  dx)=exp⁡(ln⁡(274)−1)=274e \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:

lim⁡n→∞∑i=1n1n(1+1n)1+in\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→∞n\rightarrow \infty, 1+1/n=11+1/n=1 and the rest of the expression is same as:

∫0111+xdx=2(2−1)\displaystyle \int_0^1 \frac{1}{\sqrt{1+x}}dx=2(\sqrt{2}-1)

Pranav Arora - 7 years, 5 months ago

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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 - 7 years, 5 months ago

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Thank you. :)

I eagerly wait for your upcoming posts.

Pranav Arora - 7 years, 5 months ago
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