Playing with Integrals: Limits and Integrals 1

As I said in previous post the formal definition of the Riemann Integral is very useful when solving Olympiad problems.

Problem 1. Find the following limit limn(1n+1+1n+2+...+12n)\lim_{n\to\infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}\right)

Solution. Let try to make our sum into something more or less similar to Riemann sum. i=1n1n+i=1n(i=1n11+in).\sum_{i=1}^n\frac{1}{n+i}=\frac{1}{n}\left(\sum_{i=1}^n\frac{1}{1+\frac{i}{n}}\right).

Does this remind you of the monstrous i=1nf(ξi)(xixi1)\displaystyle\sum_{i=1}^n f(\xi_i)(x_i-x_{i-1})? But what if set xi=inx_i=\dfrac{i}{n} and consider the right Riemann sum? Now it will transform into 1n(i=1nf(in))\dfrac{1}{n}\left(\displaystyle\sum_{i=1}^n f\left(\dfrac{i}{n}\right)\right).

From the last formula we can easily understand what function ff we need to consider, so by the definition: limn(1n+1+1n+2+...+12n)=1n(i=1n11+in)=\lim_{n\to\infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}\right)=\frac{1}{n}\left(\sum_{i=1}^n\frac{1}{1+\frac{i}{n}}\right)= =1n(i=1nf(in))=011x+1=ln2.=\dfrac{1}{n}\left(\displaystyle\sum_{i=1}^n f\left(\dfrac{i}{n}\right)\right)=\int^1_0\frac{1}{x+1}=\boxed{\ln 2}.

Now using the same approach try solving the following problems.

Problem 2. limn(1n+1+1n+2+...+14n)\lim_{n\to\infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{4n}\right)

Problem 3. limn(nn2+1+nn2+4+...+nn2+n2)\lim_{n\to\infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+...+\frac{n}{n^2+n^2}\right)

Note by Nicolae Sapoval
5 years, 8 months ago

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Problem 2 is same as Problem 1, the only difference is that the integration limits change. The lower limit is 0 and upper limit is 3. Hence, the answer is ln4\ln 4.

Problem 3 can be written as:

limn1n(r=1n11+(rn)2)\displaystyle \lim_{n\rightarrow \infty} \frac{1}{n}\left(\sum_{r=1}^n \cfrac{1}{1+\left(\frac{r}{n}\right)^2} \right)

The above is equivalent to

01dx1+x2=π4\displaystyle \int_{0}^1 \frac{dx}{1+x^2}=\frac{\pi}{4}

Pranav Arora - 5 years, 8 months ago

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Good job!

Nicolae Sapoval - 5 years, 8 months ago

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bt i dnt undrstand how it is 1 + x^2 & in the problem 1 only x??

Jncy Rana - 5 years, 8 months ago

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Because to get the answer we have to consider a variable (say x) Which will be in form of r/n after modifying the problem.

Manish Mishra - 5 years, 8 months ago

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Is this from IMC ?

Weijie Chen - 5 years, 8 months ago

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Sorry but I didn't get your question. Can you please explain what is "IMC"?

Pranav Arora - 5 years, 8 months ago

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@Pranav Arora It's International Mathematical Competition and it's the IMO of university studens

Weijie Chen - 5 years, 8 months ago

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in problem 3. how we can put the limits from 0 to 3??????????

Attaullah Khan - 5 years, 7 months ago

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