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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
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