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Question Number 41705 by abdo.msup.com last updated on 11/Aug/18

find nsture of the serie Σ u_n with u_n =^n (√(n/(n+1))) −1

findnstureoftheserieΣunwithun=nnn+1−1

Commented by maxmathsup by imad last updated on 12/Aug/18

we have u_n = (1−(1/(n+1)))^(1/n) −1 = e^((1/n)ln(1−(1/(n+1)))) −1 but we have ln^′ (1−u) =((−1)/(1−u)) =−Σ_(n=0) ^∞ u^n ⇒ln(1−u) =−Σ_(n=0) ^∞ (u^(n+1) /(n+1)) =− Σ_(n=1) ^∞ (u^n /n) ⇒ln(1−u) =−u −(u^2 /2) +o(u^3 ) ⇒ ln(1−(1/(n+1))) =−(1/(n+1)) −(1/(2(n+1)^2 )) +o((1/((n+1)^3 ))) ⇒ (1/n)ln(1−(1/(n+1))) =−(1/(n(n+1))) −(1/(2n(n+1)^2 )) +o((1/n^2 )) ⇒ u_n ∼ e^(−(1/(n(n+1)))) −1 ∼ 1−(1/(n(n+1))) −1 ⇒u_n ∼ −(1/(n(n+1))) but the serie Σ (1/(n(n+1))) converges ⇒ Σ u_n fonverges.

wehaveun=(1−1n+1)1n−1=e1nln(1−1n+1)−1butwehaveln′(1−u)=−11−u=−∑n=0∞un⇒ln(1−u)=−∑n=0∞un+1n+1=−∑n=1∞unn⇒ln(1−u)=−u−u22+o(u3)⇒ln(1−1n+1)=−1n+1−12(n+1)2+o(1(n+1)3)⇒1nln(1−1n+1)=−1n(n+1)−12n(n+1)2+o(1n2)⇒un∼e−1n(n+1)−1∼1−1n(n+1)−1⇒un∼−1n(n+1)buttheserieΣ1n(n+1)converges⇒Σunfonverges.

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