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Question Number 101770 by john santu last updated on 04/Jul/20

lim_(n→∞)  ((1^(13) +2^(13) +3^(13) +4^(13) +...+n^(13) )/n^(14) ) ?

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}^{\mathrm{13}} +\mathrm{2}^{\mathrm{13}} +\mathrm{3}^{\mathrm{13}} +\mathrm{4}^{\mathrm{13}} +...+{n}^{\mathrm{13}} }{{n}^{\mathrm{14}} }\:? \\ $$

Commented by Dwaipayan Shikari last updated on 04/Jul/20

(1/(14))

$$\frac{\mathrm{1}}{\mathrm{14}} \\ $$

Commented by john santu last updated on 04/Jul/20

yes. n→∞. typo.

$${yes}.\:{n}\rightarrow\infty.\:{typo}. \\ $$

Answered by bemath last updated on 05/Jul/20

lim_(n→∞)  (1/n)Σ_(i = 1) ^(13) ((i/n))^(13) = ∫_0 ^1  x^(13)  dx   = (1/(14)) .★

$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{i}\:=\:\mathrm{1}} {\overset{\mathrm{13}} {\sum}}\left(\frac{{i}}{{n}}\right)^{\mathrm{13}} =\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:{x}^{\mathrm{13}} \:{dx}\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{14}}\:.\bigstar\: \\ $$

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