Question

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

Answer 1

`1/(a+1)`.

Explanation:

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.!