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Question Number 141163 by Niiicooooo last updated on 16/May/21

Answered by mindispower last updated on 16/May/21

ζ(n)=1+Σ_(k≥2) (1/k^n ) k^n ≥nk^2 easy to see that ,k≥2 for n>>2 ⇔k^(n−2) ≥n..(E) by induction n=4,k^2 ≥4,2^2 ≥4 true suppose ∀n≥4 ..(E) true k^(n−1) =k(k^(n−2) )≥k.n≥2n≥n+1, so ∀k≥2,n≥4 k^n ≥nk^2 ⇒ 1≤ζ(n)=1+Σ_(k≥2) (1/k^n )≤1+Σ_(k≥2) (1/(nk^2 ))=1+(1/n)(ζ(2)−1) ⇒1≤lim_(n→∞) ζ(n)≤lim_(n→∞) (1+(1/n)(ζ(2)−1)) ⇔lim_(n→∞) ζ(n)=1

ζ(n)=1+∑k⩾21knkn⩾nk2easytoseethat,k⩾2forn>>2⇔kn−2⩾n..(E)byinductionn=4,k2⩾4,22⩾4truesuppose∀n⩾4..(E)truekn−1=k(kn−2)⩾k.n⩾2n⩾n+1,so∀k⩾2,n⩾4kn⩾nk2⇒1⩽ζ(n)=1+∑k⩾21kn⩽1+∑k⩾21nk2=1+1n(ζ(2)−1)⇒1⩽limn→∞ζ(n)⩽limn→∞(1+1n(ζ(2)−1))⇔limn→∞ζ(n)=1

Commented by Niiicooooo last updated on 17/May/21

greatful

greatful

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