Menu Close

study-the-convergence-of-u-n-k-1-n-1-k-1-C-n-k-k-for-that-use-H-n-k-1-n-1-k-

Tinku Tara 0 Comments
Pin




Question Number 30214 by abdo imad last updated on 18/Feb/18
study the convergence of u_n = Σ_(k=1) ^n (−1)^(k−1) (C_n ^k /k) for that use H_n = Σ_(k=1) ^n (1/k) .
studytheconvergenceofun=∑k=1n(−1)k−1CnkkforthatuseHn=∑k=1n1k.
Commented by prof Abdo imad last updated on 22/Feb/18
let condider p(x)= Σ_(k=1) ^n (−1)^(k−1) (C_n ^k /k) x^k we have p^′ (x)= Σ_(k=1) ^n C_n ^k (−1)^(k−1) x^(k−1) =((−1)/x)Σ_(k=1) ^n C_n ^k (−1)^k x^k =−(1/x)(Σ_(k=0) ^n (−1)^k x^k −1) =(1/x)( 1−(1−x)^n )=((1−(1−x)^n )/x) ⇒ p(x)= ∫_0 ^x ((1−(1−t)^n )/t)dt +λ but λ=p(0)=0⇒ p(x)= ∫_0 ^x ((1−(1−t)^n )/t)dt and u_n =p(1)⇒ u_n = ∫_0 ^1 ((1−(1−t)^n )/t)dt = = ∫_0 ^1 ((1+(1−t) +(1−t)^2 +....(1−t)^(n−1) )dt =∫_0 ^1 Σ_(k=0) ^(n−1) (1−t)^k dt= Σ_(k=0) ^(n−1) ∫_0 ^1 (1−t)^k dt =Σ_(k=0) ^(n−1) [((−1)/(k+1))(1−t)^(k+1) ]_0 ^1 =Σ_(k=0) ^(n−1) (1/(k+1)) = H_n but H_n ∼ln(n) for n→∞ ⇒lim_(n→∞) u_n =+∞.
letcondiderp(x)=∑k=1n(−1)k−1Cnkkxkwehavep′(x)=∑k=1nCnk(−1)k−1xk−1=−1x∑k=1nCnk(−1)kxk=−1x(∑k=0n(−1)kxk−1)=1x(1−(1−x)n)=1−(1−x)nx⇒p(x)=∫0x1−(1−t)ntdt+λbutλ=p(0)=0⇒p(x)=∫0x1−(1−t)ntdtandun=p(1)⇒un=∫011−(1−t)ntdt==∫01((1+(1−t)+(1−t)2+….(1−t)n−1)dt=∫01∑k=0n−1(1−t)kdt=∑k=0n−1∫01(1−t)kdt=∑k=0n−1[−1k+1(1−t)k+1]01=∑k=0n−11k+1=HnbutHn∼ln(n)forn→∞⇒limn→∞un=+∞.

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *