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 limn(1nln(n!)lnn)\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. limn(1nln(n!)lnn)=limn1n(ln(n!)nlnn)=\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)= =limn1n(i=1nln(in))=01lnxdx.=\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 01lnxdx=(xlnxx)01=1+limx0(xlnx)=1.\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 limn(sin1n2+1+sin1n2+4+...+sin110n2)\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.

limn(i=13nsin1n2+i2)=limn(i=13nsin1n1+i2n2).\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\sin function. Many of you proably know that near the 00 function sinxsin x behaves just like xx. This fact is proven by computing Taylor Series of sinxsin x at the point x=0x=0.

limn(i=13nsin1n1+i2n2)=limn(i=13n1n1+i2n2)+\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(limni=13n(1n1+i2n2)3)=031x2+1dx=+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= =ln(3+10).=\boxed{\ln(3+\sqrt{10})}.

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

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

Note by Nicolae Sapoval
5 years, 9 months ago

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Well done , Nicolae :)

_Problem 9 _

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

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

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

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

= exp(01lnxdx)\displaystyle \text{exp} \bigg(\displaystyle \int_{0}^{1} \ln x \text{dx} \bigg)

We know by using by parts that lnxdx=xlnxx\displaystyle \int \ln x \text{dx} = x \ln x - x , and we also know that limx0xlnx=0\displaystyle \lim_{x \to 0} x \ln x = 0, hence, the integral 01lnxdx\displaystyle \int_{0}^{1} \ln x \text{dx} turns out to be 1-1, hence, finally, we conclude that answer to the problem is e1\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 limnr=12n(r+nn)rn2 \displaystyle \lim_{n \to \infty} \displaystyle \prod_{r=1}^{2n} \bigg(\frac{r+n}{n}\bigg)^{\frac{r}{n^2}}

jatin yadav - 5 years, 9 months ago

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Nice post again Nicolae! :)

Solution to Problem 10:

The given limit is equivalent to

limnexp(1nr=12nrnln(1+rn))\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)

= exp(02xln(1+x)dx)\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(3/2)\ln2. Hence the final answer is 222\sqrt{2}.

Pranav Arora - 5 years, 9 months ago

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Shouldn't it be 333\sqrt{3} ? As 02xln(1+x)dx=(3/2)ln3\displaystyle \int_0^{2} x\ln(1+x) \mathrm d x=(3/2)\ln3

Abhijeeth Babu - 5 years, 9 months ago

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

Pranav Arora - 5 years, 9 months ago

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@Pranav Arora No problem bro :D

Abhijeeth Babu - 5 years, 9 months ago

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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 - 5 years, 9 months ago

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Nice Job!

Bob Krueger - 5 years, 9 months ago

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Really good job bud.

Abhijeeth Babu - 5 years, 9 months ago

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