Question

Question is from definite integral.

Answer 1

Answer:

The value of the sum is: $\frac{\pi}{4}$.

Step-by-step explanation:

Given: $\lim_{n \to \infty} \displaystyle \sum_{k = 1}^ {n} \bigg ( \frac{n}{n^2 + k^2} \bigg )$.

Taking $n^2$ common outside from the denominator, we have:

$\bigg ( \frac{n}{(n^2)(1 + (\frac{k^2}{n^2})} \bigg )$

$\implies\frac{1}{n} .\frac{1}{1 +\big ( \frac {k^2}{n^2}\big )}$

We have the following theorem.

If $f$ is integrable on [0, 1] then $\int _{0}^{1} f(x) dx = \lim_{n \to \infty} \frac{1}{n} \displaystyle \sum_{r = 1}^{n} f( \frac{x}{n})$.

Now, let $\frac{k}{n} = x$

$\implies dx = \frac{1}{n}$.

Therefore the summation becomes

$= \int_{0}^{1} \bigg ( \frac{1}{1 + x^2} \bigg )$

$\implies \tan^{-1}(x)|_{0}^{1}$

$\implies tan^{-1}(1) - tan^{-1}(0)$

$= \frac{\pi}{4} - 0$

$= \frac{\pi}{4}$

Hence, the sum is $\pi$/4.