To prove the result for arbitrary real a>-1 (for integer a, it’s easier to prove it using induction), by using an approach similar to the one used for proving the theorem 10.11, I proved first that lim_{n -> infinity} [ sum(k^a) / (n+1) ] = 1/(a+1). With that, it’s somewhat obvious how to to derive the final result.
Andres Tellez says:
See exercise 13 (c)- section I 4.10
Andres Tellez says:
Is better to see exercise 34(c)- section 10.4. You can compute the limit easily from there.
Anonymous says:
Others pointed out the solutions for a >= 0 but for a < 0 it's still not clear. However, for a 0 So the whole limit is goes to 1/nC_1/C_2 which goes to 0. The interval -2 to 0 still remains.
Anonymous says:
could you please elaborate
Roberto says:
Hint: Sn(a) is asymptotically equivalent to n^(a+1)/(a+1)
To prove the result for arbitrary real a>-1 (for integer a, it’s easier to prove it using induction), by using an approach similar to the one used for proving the theorem 10.11, I proved first that lim_{n -> infinity} [ sum(k^a) / (n+1) ] = 1/(a+1). With that, it’s somewhat obvious how to to derive the final result.
See exercise 13 (c)- section I 4.10
Is better to see exercise 34(c)- section 10.4. You can compute the limit easily from there.
Others pointed out the solutions for a >= 0 but for a < 0 it's still not clear. However, for a 0 So the whole limit is goes to 1/nC_1/C_2 which goes to 0. The interval -2 to 0 still remains.
could you please elaborate
Hint: Sn(a) is asymptotically equivalent to n^(a+1)/(a+1)
If a != -1
Can you please explain further?