# 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 $\lim_{n\to\infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}\right)$

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

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

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

Now using the same approach try solving the following problems.

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

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

Note by Nicolae Sapoval
6 years, 2 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 $\ln 4$.

Problem 3 can be written as:

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

The above is equivalent to

$\displaystyle \int_{0}^1 \frac{dx}{1+x^2}=\frac{\pi}{4}$

- 6 years, 2 months ago

Good job!

- 6 years, 2 months ago

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

- 6 years, 2 months ago

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.

- 6 years, 2 months ago

Is this from IMC ?

- 6 years, 2 months ago

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

- 6 years, 2 months ago

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

- 6 years, 2 months ago

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

- 6 years, 1 month ago