Playing with Integrals: Limits and Integrals 3

Here comes the last part of the Limits and Integrals series, previous ones can be found here: Part 1 and Part 2. This time we will discuss even more challenging limits, so get ready and have fun!

The next problem was found in "A Note on Evaluating Limits Using Riemann Sums" by Sudhir K. Goel and Dennis M. Rodriguez.

Problem 7. Evaluate $\lim_{n\to\infty}\left(\frac{1}{n}\ln(n!)-\ln n\right)$

Solution. After mastering several similar problems there should be no dificulty in applying our technique to this one. $\lim_{n\to\infty}\left(\frac{1}{n}\ln(n!)-\ln n\right)=\lim_{n\to\infty}\frac{1}{n}(\ln(n!)-n\ln n)=$ $=\lim_{n\to\infty}\frac{1}{n}\left(\sum^n_{i=1}\ln\left(\frac{i}{n}\right)\right)=\int^1_0\ln x\,dx.$

Even slightest knowledge of integrals suggest that something went wrong. Actually everything is just fine, but the intgeral we need to evaluate is an improper one. Now let's proceed to finish it $\int^1_0\ln x\,dx=\left. (xlnx-x)\right|^1_0=-1+\lim_{x\to 0}(x\ln x)=-1.$

Now let's move onto the next example.

Problem 8. Calculate $\lim_{n\to\infty}\left(\sin\frac{1}{\sqrt{n^2+1}}+\sin\frac{1}{\sqrt{n^2+4}}+...+\sin\frac{1}{\sqrt{10n^2}}\right)$

Solution. This one will be a tough task, but after eight solved examples nothing is imposible.

$\lim_{n\to\infty}\left(\sum^{3n}_{i=1}\sin\frac{1}{\sqrt{n^2+i^2}}\right)=\lim_{n\to\infty}\left(\sum^{3n}_{i=1}\sin\frac{1}{n\sqrt{1+\frac{i^2}{n^2}}}\right).$

Now we need to get rid of the $\sin$ function. Many of you proably know that near the $0$ function $sin x$ behaves just like $x$ . This fact is proven by computing Taylor Series of $sin x$ at the point $x=0$ .

$\lim_{n\to\infty}\left(\sum^{3n}_{i=1}\sin\frac{1}{n\sqrt{1+\frac{i^2}{n^2}}}\right)=\lim_{n\to\infty}\left(\sum^{3n}_{i=1}\frac{1}{n\sqrt{1+\frac{i^2}{n^2}}}\right)+$ $+O\left(\lim_{n\to\infty}\sum^{3n}_{i=1}\left(\frac{1}{n\sqrt{1+\frac{i^2}{n^2}}}\right)^3\right)=\int^3_0\frac{1}{\sqrt{x^2+1}}\,dx=$ $=\boxed{\ln(3+\sqrt{10})}.$

The last problem in this series of posts we'll be following one.

Problem 9. Find the limit $\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}$

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Math	Appears 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

Well done , Nicolae :)

_Problem 9 _

Solution: $\displaystyle \lim_{n \to \infty} \frac{\sqrt[n] {n!}}{n}$

= $\displaystyle \lim_{n \to \infty} {\bigg(\frac{ {n!}}{n^n}\bigg)}^{\frac{1}{n}}$

= $\displaystyle \lim_{n \to \infty} {\bigg(\prod_{r= 1}^{n} \frac{r}{n}\bigg)}^{\frac{1}{n}}$

= $\displaystyle \lim_{n \to \infty} \text{exp}\bigg(\frac{1}{n} \sum_{r=1}^{n} \ln \bigg(\frac{r}{n}\bigg)\bigg)$

= $\displaystyle \text{exp} \bigg(\displaystyle \int_{0}^{1} \ln x \text{dx} \bigg)$

We know by using by parts that $\displaystyle \int \ln x \text{dx} = x \ln x - x$ , and we also know that $\displaystyle \lim_{x \to 0} x \ln x = 0$ , hence, the integral $\displaystyle \int_{0}^{1} \ln x \text{dx}$ turns out to be $-1$ , hence, finally, we conclude that answer to the problem is $\boxed{e^{-1}}$

I had a question abt latex, actually, i wanted to know how do we put limits on an indefinte integral , when solving a definite integral on latex.

If you don't mind , i add problem 10 here:

Problem 10 Evaluate $\displaystyle \lim_{n \to \infty} \displaystyle \prod_{r=1}^{2n} \bigg(\frac{r+n}{n}\bigg)^{\frac{r}{n^2}}$

jatin yadav - 7 years, 5 months ago

Nice post again Nicolae! :)

Solution to Problem 10:

The given limit is equivalent to

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

= $\displaystyle \exp\left(\int_0^2 x\ln(1+x)\,\text{d}x \right)$

The above integral can be easily evaluated using integration by parts and is equal to $(3/2)\ln2$ . Hence the final answer is $2\sqrt{2}$ .

Pranav Arora - 7 years, 5 months ago

Shouldn't it be $3\sqrt{3}$ ? As $\displaystyle \int_0^{2} x\ln(1+x) \mathrm d x=(3/2)\ln3$

Abhijeeth Babu - 7 years, 5 months ago

@Abhijeeth Babu – You are right, sorry about that. :)

@Pranav Arora – No problem bro :D

P.S. I hope that during the next few days I'll make/find about 4-5 problems involving this technique and will add them as problems. I also will post a note leading to all 3 posts and related problems.

Nicolae Sapoval - 7 years, 5 months ago

Nice Job!

Bob Krueger - 7 years, 5 months ago

Really good job bud.

Math	Appears 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}`	$\boxed{123}$