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let-S-n-k-1-n-1-k-k-1-calculate-S-n-interms-of-H-n-2-find-lim-n-S-n-3-let-W-n-1-i-lt-j-n-1-i-j-i-j-prove-that-W-n-is-convergent-and-calculste-its-lim

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Question Number 40067 by abdo mathsup 649 cc last updated on 15/Jul/18
let S_n = Σ_(k=1) ^n (((−1)^k )/k) 1) calculate S_n interms of H_n 2) find lim_(n→+∞) S_n 3) let W_n = Σ_(1≤i<j≤n) (((−1)^(i+j) )/(i.j)) prove that (W_n ) is convergent and calculste its limit.
letSn=∑k=1n(−1)kk1)calculateSnintermsofHn2)findlimn→+∞Sn3)letWn=∑1⩽i<j⩽n(−1)i+ji.jprovethat(Wn)isconvergentandcalculsteitslimit.
Commented by abdo mathsup 649 cc last updated on 17/Jul/18
S_n = Σ_(k=1and k=2p) ^n (((−1)^k )/k) +Σ_(k=1 and k=2p+1) ^n (((−1)^k )/k) =Σ_(p=1) ^([(n/2)]) (1/(2p)) − Σ_(p=0) ^([((n−1)/2)]) (1/(2p+1)) but Σ_(p=1) ^([(n/2)]) (1/(2p)) =(1/2) H_([(n/2)]) Σ_(p=0) ^([((n−1)/2)]) (1/(2p+1)) =1 +(1/3) + (1/5) +....+(1/(2[((n−1)/2)]+1)) =1+(1/2) +(1/3) +(1/4) +(1/5) +.....+(1/(2[((n−1)/2)])) +(1/(2[((n−1)/2)]+1)) −(1/2) −(1/4) −....−(1/(2[((n−1)/2)])) =H_(2[((n−1)/2)]+1) −(1/2) H_([((n−1)/2)]) ⇒ S_n = (1/2) H_([(n/2)]) −H_(2[((n−1)/2)]+1) +(1/2) H_([((n−1)/2)]) 2) we have S_(2n) =(1/2) H_n +(1/2) H_(n−1) −H_(2n−1) S_(2n) =(1/2){ ln(n) +γ +ln(n−1) +γ +o((1/n))} −ln(2n−1)−γ +o((1/n)) =(1/2)ln(n^2 −n)−ln(2n−1) +o((1/n)) =ln(((√(n^2 −n))/(2n−1))) +o((1/n)) ⇒lim_(n→+∞) S_(2n) = −ln(2) also S_(2n+1) =(1/2) H_n −H_(2n+1) +(1/2) H_n = H_n −H_(2n+1) = ln(n) +γ +o((1/n)) −ln(2n+1)−γ −o((1/n)) =ln((n/(2n+1))) +o((1/n)) ⇒lim_(n→+∞) S_(2n+1) = −ln(2) from that we can conclude that lim_(n→+∞) S_n =−ln(2).
Sn=∑k=1andk=2pn(−1)kk+∑k=1andk=2p+1n(−1)kk=∑p=1[n2]12p−∑p=0[n−12]12p+1but∑p=1[n2]12p=12H[n2]∑p=0[n−12]12p+1=1+13+15+….+12[n−12]+1=1+12+13+14+15+…..+12[n−12]+12[n−12]+1−12−14−….−12[n−12]=H2[n−12]+1−12H[n−12]⇒Sn=12H[n2]−H2[n−12]+1+12H[n−12]2)wehaveS2n=12Hn+12Hn−1−H2n−1S2n=12{ln(n)+γ+ln(n−1)+γ+o(1n)}−ln(2n−1)−γ+o(1n)=12ln(n2−n)−ln(2n−1)+o(1n)=ln(n2−n2n−1)+o(1n)⇒limn→+∞S2n=−ln(2)alsoS2n+1=12Hn−H2n+1+12Hn=Hn−H2n+1=ln(n)+γ+o(1n)−ln(2n+1)−γ−o(1n)=ln(n2n+1)+o(1n)⇒limn→+∞S2n+1=−ln(2)fromthatwecanconcludethatlimn→+∞Sn=−ln(2).
Commented by abdo mathsup 649 cc last updated on 17/Jul/18
3) we have S_n ^2 =(Σ_(k=1) ^n (((−1)^k )/k))^2 =Σ_(k=1) ^n (1/k^2 ) +2 Σ_(1≤i<j≤n) (((−1)^i (−1)^j )/(i.j)) =Σ_(k=1) ^n (1/k^2 ) +2W_n ⇒ 2W_n =S_n ^2 −Σ_(k=1) ^n (1/k^2 ) ⇒ lim_(n→+∞) W_n = (1/2)( (−ln2)^2 −(π^2 /6)) =(1/2)( (ln(2))^2 −(π^2 /6)) .
3)wehaveSn2=(∑k=1n(−1)kk)2=∑k=1n1k2+2∑1⩽i<j⩽n(−1)i(−1)ji.j=∑k=1n1k2+2Wn⇒2Wn=Sn2−∑k=1n1k2⇒limn→+∞Wn=12((−ln2)2−π26)=12((ln(2))2−π26).

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