1Thanks sir... i got it! :)
\lim_{n->infinity}(\sum_{r=1}^{m}{r^{n}})^{1/n} is equal to: (n belongs to Natural nos.)
a) m b) m/2
c)em d)em/2
Ans: (a)
please tell me the method to solve this.....
(SOURCE ARIHANT DIFF CALC.)
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2 Answers
1708\hspace{-16}\bf{\lim_{n\rightarrow \infty} \sum_{r=1}^{m}(r^n)^{\frac{1}{n}}=\lim_{n\rightarrow \infty}\{1^n+2^n+3^n+.....+m^n\}^{\frac{1}{n}}}\\\\\\ \bf{=\lim_{n\rightarrow \infty} m.\left\{\left(\frac{1}{m}\right)^n+\left(\frac{2}{m}\right)^n+.......+\left(\frac{m-1}{m}\right)^n+\left(\frac{m}{m}\right)^n\right\}^{\frac{1}{n}}}$\\\\\\ So $\bf{\lim_{n\rightarrow \infty} \sum_{r=1}^{m}(r^n)^{\frac{1}{n}}=m}$