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  4. Let Sn=n-2015(1+22014+32014+ ldots ldots .+n2014), then undersetn arrow ∞ textLim Sn is

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Q. Let $S_n=n^{-2015}\left(1+2^{2014}+3^{2014}+\ldots \ldots .+n^{2014}\right)$, then $\underset{n \rightarrow \infty}{\text{Lim}} S_n$ is

Integrals

Solution:

$S _{ n }=\displaystyle\sum_{ i =1}^{ n } \frac{1}{ n }\left(\frac{ i }{ n }\right)^{2014}=\int\limits_0^1 x ^{2014} dx =\frac{1}{2015}$.