N -lim1P - 2P - 3P - - nPnP- 1 equals-

n →∞lim1^P + 2^P + 3^P + .... +n^Pn^P+ 1 =n →∞lim1n1^P + 2^P + 3^P + .... +n^Pn^P =n →∞lim1n( 1^Pn^p + 2^Pn^p + 3^Pn^P + .... +n^Pn^P) =n →∞ ...

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Summation by Sigma Method

n →∞lim1P + 2...

Question

$l i m n \to \infty \frac{1^{P} + 2^{P} + 3^{P} + . . . . + n^{P}}{n^{P + 1}}$ equals-

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Solution

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