#lim_(n->oo)(1^alpha+2^alpha+cdots+n^alpha)/n^(alpha+1) =#?

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Answer 1 · 2017-04-27T13:43:05

#1/(a+1)#.

Recall that, #int_a^b f(x)dx=lim_(n to oo) sum_(r=1)^(r=n)hf(a+rh),#

where, #h=(b-a)/n.#

If we take, #a=0, and, b=1," so that, "h=1/n,# we have,

#int_0^1 f(x)dx=lim_(n to oo) sum_(r=1)^(r=n)1/n*f(r/n).......(star).#

Now, the Reqd. Lim.=#lim_(n to oo) (1^a+2^a+3^a+...+n^a)/n^(a+1)#

#=lim_(n to oo) [1/n{(1/n)^a+(2/n)^a+(3/n)^a+...+(n/n)^a}]#

#=lim_(n to oo) [sum_(r=1)^(r=n)1/n*(r/n)^a].#

#:., (star) rArr" The Limit="int_0^1 x^adx,#

#=[x^(a+1)/(a+1)]_0^1,#

#"The Limit="1/(a+1).#

Enjoy Maths.!