lim n~ infinity (1^k+ 2^k+ 3^k+.........+n^k)\ (k*(n^k + 1)

More generally, we can say:
[1.5]

ALSO

PUTIN K=1 WE GET

LIM = (N^2 + N)/2(N+1)=1/2*1/0=INFINITY

PUTING N=K WE GET INFINITY

PINT N=K+1 WE GET iNFINITY

tHEREFORTH sINCE THERE ARE GREATER POWERS OF N AT NUMERRATOR W.R.T dINOMINATOR THERE FORTH tHE ANSWER IS INFINITY