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Question Number 30175 by abdo imad last updated on 17/Feb/18

prove that u_n = Σ_(k=1) ^n (1/(n+k)) is convergente .

provethatun=∑k=1n1n+kisconvergente.

Commented by abdo imad last updated on 21/Feb/18

we have u_n = (1/n) Σ_(k=1) ^n (1/(1 +(k/n))) ⇒ lim_(n→∞) u_n =lim_(n→∞) ((1−0)/n)Σ_(k=1) ^n (1/(1+((k(1−0))/n))) (Rieman sum) = ∫_0 ^1 (dx/(1+x)) =[ln∣1+x∣]_0 ^1 = ln2 .

wehaveun=1n∑k=1n11+kn⇒limn→∞un=limn→∞1−0n∑k=1n11+k(1−0)n(Riemansum)=∫01dx1+x=[ln∣1+x∣]01=ln2.

Commented by abdo imad last updated on 21/Feb/18

another method we have u_n = (1/(n+1)) +(1/(n+2)) +.... (1/(2n))=1 +(1/2) +(1/3) +...+(1/n) +(1/(n+1)) +... +(1/(2n)) −(1+(1/2) +(1/3) +...+(1/n))=H_(2n) − H_n but H_(2n) =ln(2n) +γ +o((1/n)) and H_n = ln(n) +γ +o((1/n))⇒ H_(2n) −H_n =ln(((2n)/n)) +o((1/n)) ⇒lim_(n→∞) H_(2n) −H_n =ln(2).so lim_(n→∞) u_n =ln(2) .

anothermethodwehaveun=1n+1+1n+2+....12n=1+12+13+...+1n+1n+1+...+12n−(1+12+13+...+1n)=H2n−HnbutH2n=ln(2n)+γ+o(1n)andHn=ln(n)+γ+o(1n)⇒H2n−Hn=ln(2nn)+o(1n)⇒limn→∞H2n−Hn=ln(2).solimn→∞un=ln(2).

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