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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.
$${let}\:\:{S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{S}_{{n}} \:{interms}\:{of}\:{H}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:{let}\:{W}_{{n}} =\:\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \frac{\left(−\mathrm{1}\right)^{{i}+{j}} }{{i}.{j}} \\ $$$${prove}\:{that}\:\left({W}_{{n}} \right)\:{is}\:{convergent}\:{and}\:{calculste}\:{its} \\ $$$${limit}. \\ $$
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).
$${S}_{{n}} =\:\sum_{{k}=\mathrm{1}{and}\:{k}=\mathrm{2}{p}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\:\:\:+\sum_{{k}=\mathrm{1}\:{and}\:{k}=\mathrm{2}{p}+\mathrm{1}} ^{{n}} \:\:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}} \\ $$$$=\sum_{{p}=\mathrm{1}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:\:\frac{\mathrm{1}}{\mathrm{2}{p}}\:\:\:\:−\:\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:\:\frac{\mathrm{1}}{\mathrm{2}{p}+\mathrm{1}}\:\:{but}\: \\ $$$$\sum_{{p}=\mathrm{1}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:\frac{\mathrm{1}}{\mathrm{2}{p}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{\left[\frac{{n}}{\mathrm{2}}\right]} \\ $$$$\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:\frac{\mathrm{1}}{\mathrm{2}{p}+\mathrm{1}}\:=\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{3}}\:\:+\:\frac{\mathrm{1}}{\mathrm{5}}\:+….+\frac{\mathrm{1}}{\mathrm{2}\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]+\mathrm{1}} \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{5}}\:+…..+\frac{\mathrm{1}}{\mathrm{2}\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]}\:+\frac{\mathrm{1}}{\mathrm{2}\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]+\mathrm{1}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{4}}\:−….−\frac{\mathrm{1}}{\mathrm{2}\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \\ $$$$={H}_{\mathrm{2}\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]+\mathrm{1}} \:\:−\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:\:\:\:\Rightarrow \\ $$$${S}_{{n}} =\:\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{\left[\frac{{n}}{\mathrm{2}}\right]} \:\:−{H}_{\mathrm{2}\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]+\mathrm{1}} \:\:+\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{S}_{\mathrm{2}{n}} \:=\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}} \:\:+\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}−\mathrm{1}} \:\:−{H}_{\mathrm{2}{n}−\mathrm{1}} \\ $$$${S}_{\mathrm{2}{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\:{ln}\left({n}\right)\:+\gamma\:\:+{ln}\left({n}−\mathrm{1}\right)\:+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\right\} \\ $$$$−{ln}\left(\mathrm{2}{n}−\mathrm{1}\right)−\gamma\:\:+{o}\left(\frac{\mathrm{1}}{{n}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({n}^{\mathrm{2}} −{n}\right)−{ln}\left(\mathrm{2}{n}−\mathrm{1}\right)\:+{o}\left(\frac{\mathrm{1}}{{n}}\right) \\ $$$$={ln}\left(\frac{\sqrt{{n}^{\mathrm{2}} −{n}}}{\mathrm{2}{n}−\mathrm{1}}\right)\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:\:{S}_{\mathrm{2}{n}} \:=\:−{ln}\left(\mathrm{2}\right) \\ $$$${also} \\ $$$${S}_{\mathrm{2}{n}+\mathrm{1}} \:\:=\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}} \:\:\:−{H}_{\mathrm{2}{n}+\mathrm{1}} \:\:+\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}} \\ $$$$=\:{H}_{{n}} \:−{H}_{\mathrm{2}{n}+\mathrm{1}} =\:{ln}\left({n}\right)\:+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:−{ln}\left(\mathrm{2}{n}+\mathrm{1}\right)−\gamma\:−{o}\left(\frac{\mathrm{1}}{{n}}\right) \\ $$$$={ln}\left(\frac{{n}}{\mathrm{2}{n}+\mathrm{1}}\right)\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:\:{S}_{\mathrm{2}{n}+\mathrm{1}} \:=\:−{ln}\left(\mathrm{2}\right) \\ $$$${from}\:{that}\:{we}\:{can}\:{conclude}\:{that}\: \\ $$$${lim}_{{n}\rightarrow+\infty} \:\:{S}_{{n}} =−{ln}\left(\mathrm{2}\right). \\ $$$$ \\ $$
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)) .
$$\left.\mathrm{3}\right)\:{we}\:{have}\:{S}_{{n}} ^{\mathrm{2}} \:=\left(\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\right)^{\mathrm{2}} \\ $$$$=\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:\:\:+\mathrm{2}\:\:\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{i}} \left(−\mathrm{1}\right)^{{j}} }{{i}.{j}} \\ $$$$=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:+\mathrm{2}{W}_{{n}} \:\:\Rightarrow\:\mathrm{2}{W}_{{n}} ={S}_{{n}} ^{\mathrm{2}} \:\:−\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} } \\ $$$$\Rightarrow\:{lim}_{{n}\rightarrow+\infty} \:{W}_{{n}} =\:\frac{\mathrm{1}}{\mathrm{2}}\left(\:\left(−{ln}\mathrm{2}\right)^{\mathrm{2}} \:−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\:\left({ln}\left(\mathrm{2}\right)\right)^{\mathrm{2}} \:−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\right)\:. \\ $$

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