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# 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
2 years, 10 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

$$\displaystyle \int_{0}^1 \frac{dx}{1+x^2}=\frac{\pi}{4}$$ · 2 years, 10 months ago

bt i dnt undrstand how it is 1 + x^2 & in the problem 1 only x?? · 2 years, 10 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. · 2 years, 10 months ago

Good job! · 2 years, 10 months ago

Is this from IMC ? · 2 years, 10 months ago

Sorry but I didn't get your question. Can you please explain what is "IMC"? · 2 years, 10 months ago

It's International Mathematical Competition and it's the IMO of university studens · 2 years, 10 months ago