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Compute limn??n?i=1k[f(a+in)?f(a)]
limn??n?i=1k[f(a+in)?f(a)]=?i=1k[limn??n[f(a+in)?f(a)]]
Let t=in?n=it?i=nt
?i=1k[limn??n[f(a+in)?f(a)]]=?i=1klimt?0i?f(a+t)?f(a)t=?i=1kif?(a)=k(k+1)2?f?(a)
Therefore, A=K(K+1)2?f?(a)
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Deduce the limit limn??n?i=1k[(n+i)mnm?1?1]
limn??n?i=1k[(n+i)mnm?1?1]=?i=1k[limn??n((n+i)mnm?1)]
We have: limn??n?(n+i)m?nmnm=limn??n?[(n+in)m?1]=limn??n?[(1+in)m?1]
Let u=in,n??,u?0
limn??n?[(1+in)m?1]=limu?0iu[(1+u)m?1]=limu?0i[(1+u)m?1u],bylimu?0(1+u)m?1u=m??i=1kim=mk(k+1)2
Therefore, limn??n?i=1k[(n+i)mnm?1?1]=mk(k+1)2
b).
Therefore, limn??n?i=1k[(n+i)mnm?1?1]=mk(k+1)2
a).
Therefore, A=K(K+1)2?f?(a)
