Answer:
The value of the sum is: [tex]$ \frac{\pi}{4} $[/tex].
Step-by-step explanation:
Given: [tex]$ \lim_{n \to \infty} \displaystyle \sum_{k = 1}^ {n} \bigg ( \frac{n}{n^2 + k^2} \bigg )[/tex].
Taking [tex] $ n^2 $[/tex] common outside from the denominator, we have:
[tex]$ \bigg ( \frac{n}{(n^2)(1 + (\frac{k^2}{n^2})} \bigg )[/tex]
[tex]\implies\frac{1}{n} .\frac{1}{1 +\big ( \frac {k^2}{n^2}\big )}[/tex]
We have the following theorem.
If [tex]$f $[/tex] is integrable on [0, 1] then [tex]$ \int _{0}^{1} f(x) dx = \lim_{n \to \infty} \frac{1}{n} \displaystyle \sum_{r = 1}^{n} f( \frac{x}{n}) $[/tex].
Now, let [tex]$ \frac{k}{n} = x $[/tex]
[tex]$ \implies dx = \frac{1}{n} $[/tex].
Therefore the summation becomes
[tex]$ = \int_{0}^{1} \bigg ( \frac{1}{1 + x^2} \bigg )[/tex]
[tex]$ \implies \tan^{-1}(x)|_{0}^{1} $[/tex]
[tex]$ \implies tan^{-1}(1) - tan^{-1}(0) $[/tex]
[tex]$ = \frac{\pi}{4} - 0 $[/tex]
[tex]$ = \frac{\pi}{4} $[/tex]
Hence, the sum is [tex]$ \pi $[/tex]/4.
