MCQ
$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{k}{{{n^2} + {k^2}}}} $is equals to
- ✓$\frac{1}{2}\log 2$
- B$log\ 2$
- C$\pi /4$
- D$\pi /2$
$ = \mathop {\lim }\limits_{n \to \infty } \,\sum\limits_{k = 1}^n {} \frac{1}{n}\frac{{\left( {\frac{k}{n}} \right)}}{{1 + {{\left( {\frac{k}{n}} \right)}^2}}}$
$I = \int\limits_0^1 {\frac{x}{{1 + {x^2}}}dx} $
$ = \frac{1}{2}[\log (1 + {x^2})]_{\,0}^{\,1}$$ = \frac{1}{2}\left[ {\log 2} \right]$.
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$f(n)=n+\frac{16+5 n-3 n^2}{4 n+3 n^2}+\frac{32+n-3 n^2}{8 n+3 n^2}+\frac{48-3 n-3 n^2}{12 n+3 n^2}+\ldots+\frac{25 n-7 n^2}{7 n^2}$
Then, the value of $\lim _{ n \rightarrow \infty} f( n )$ is equal to