Question

Assume that f is differentiable at a.

1. Compute limn→∞n∑i=1k[f(a+in)−f(a)]

2. Deduce the limit limn→∞n∑i=1k[(n+i)mnm−1−1]

Answer 1

• 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)

• 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)