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Question Number 45792 by maxmathsup by imad last updated on 16/Oct/18
let u_n =Σ_(k=1) ^n (((−1)^([k]) )/k) and H_n =Σ_(k=1) ^n (1/k) 1)calculate u_n interms of H_n 2)study the convergence of (u_n ) 3)study theconvergence of Σ u_(n.)
$${let}\:{u}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{\left[{k}\right]} }{{k}}\:\:{and}\:{H}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}} \\ $$$$\left.\mathrm{1}\right){calculate}\:{u}_{{n}} \:{interms}\:{of}\:{H}_{{n}} \\ $$$$\left.\mathrm{2}\right){study}\:{the}\:{convergence}\:{of}\:\left({u}_{{n}} \right) \\ $$$$\left.\mathrm{3}\right){study}\:{theconvergence}\:{of}\:\Sigma\:{u}_{{n}.} \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 17/Oct/18
3) lim_(n→+∞) u_n ≠0 ⇒ Σ u_n diverges..
$$\left.\mathrm{3}\right)\:{lim}_{{n}\rightarrow+\infty} {u}_{{n}} \neq\mathrm{0}\:\Rightarrow\:\Sigma\:{u}_{{n}} \:\:{diverges}.. \\ $$
Commented by maxmathsup by imad last updated on 17/Oct/18
2) convergence of (u_n )? we have u_(2n) =(1/2) H_n +(1/2) H_(n−1) −H_(2n−1) =(1/2)( ln(n)+γ +o((1/n)))+(1/2)(ln(n−1) +γ +o((1/n)))−ln(2n−1)−γ −o((1/n)) =ln(√(n(n+1)))−ln(2n−1) +o((1/n))=ln(((√(n^2 +n))/(2n−1)))+o((1/n))→−ln(2)(n→+∞) u_(2n+1) =(1/2) H_n +(1/2) H_n −H_(2n+1) =H_n −H_(2n+1) =ln(n) +γ −ln(2n+1)−γ +o((1/n))=ln((n/(2n+1)))+o((1/n))→−ln(2)(n→+∞) so (u_n ) converges and lim_(n→+∞) u_n =−ln(2).
$$\left.\mathrm{2}\right)\:{convergence}\:{of}\:\left({u}_{{n}} \right)? \\ $$$${we}\:{have}\:{u}_{\mathrm{2}{n}} =\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}} +\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}−\mathrm{1}} \:−{H}_{\mathrm{2}{n}−\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\:{ln}\left({n}\right)+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\right)+\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left({n}−\mathrm{1}\right)\:+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\right)−{ln}\left(\mathrm{2}{n}−\mathrm{1}\right)−\gamma\:−{o}\left(\frac{\mathrm{1}}{{n}}\right) \\ $$$$={ln}\sqrt{{n}\left({n}+\mathrm{1}\right)}−{ln}\left(\mathrm{2}{n}−\mathrm{1}\right)\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)={ln}\left(\frac{\sqrt{{n}^{\mathrm{2}} +{n}}}{\mathrm{2}{n}−\mathrm{1}}\right)+{o}\left(\frac{\mathrm{1}}{{n}}\right)\rightarrow−{ln}\left(\mathrm{2}\right)\left({n}\rightarrow+\infty\right) \\ $$$${u}_{\mathrm{2}{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}} \:\:+\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}} \:\:\:−{H}_{\mathrm{2}{n}+\mathrm{1}} ={H}_{{n}} −{H}_{\mathrm{2}{n}+\mathrm{1}} \\ $$$$={ln}\left({n}\right)\:+\gamma\:−{ln}\left(\mathrm{2}{n}+\mathrm{1}\right)−\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)={ln}\left(\frac{{n}}{\mathrm{2}{n}+\mathrm{1}}\right)+{o}\left(\frac{\mathrm{1}}{{n}}\right)\rightarrow−{ln}\left(\mathrm{2}\right)\left({n}\rightarrow+\infty\right) \\ $$$${so}\:\left({u}_{{n}} \right)\:{converges}\:{and}\:{lim}_{{n}\rightarrow+\infty} {u}_{{n}} =−{ln}\left(\mathrm{2}\right). \\ $$
Commented by maxmathsup by imad last updated on 17/Oct/18
we have u_n =Σ_(k=1) ^n (((−1)^([k]) )/k) ⇒u_n =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/(2[((n−1)/2)])) +(1/(2[((n−1)/2)]+1)) −(1/2)(1+(1/2) +....+(1/(2[((n−1)/2)]))) =H_(2[((n−1)/2)]+1) −(1/2) H_([((n−1)/2)]) ⇒ u_n =(1/2) H_([(n/2)]) +(1/2) H_([((n−1)/2)]) − H_(2[((n−1)/2)]+1)
$${we}\:{have}\:{u}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{\left[{k}\right]} }{{k}}\:\Rightarrow{u}_{{n}} =\sum_{{p}=\mathrm{1}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:\frac{\mathrm{1}}{\mathrm{2}{p}}\:−\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:\frac{\mathrm{1}}{\mathrm{2}{p}+\mathrm{1}}\:{but} \\ $$$$\sum_{{p}=\mathrm{1}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:\frac{\mathrm{1}}{\mathrm{2}{p}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{\left[\frac{{n}}{\mathrm{2}}\right]} \\ $$$$\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:\frac{\mathrm{1}}{\mathrm{2}{p}+\mathrm{1}}\:=\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{5}}\:+….+\frac{\mathrm{1}}{\mathrm{2}\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]+\mathrm{1}} \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:+…..+\frac{\mathrm{1}}{\mathrm{2}\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]}\:+\frac{\mathrm{1}}{\mathrm{2}\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]+\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:+….+\frac{\mathrm{1}}{\mathrm{2}\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]}\right) \\ $$$$={H}_{\mathrm{2}\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]+\mathrm{1}} \:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:\:\Rightarrow \\ $$$${u}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{\left[\frac{{n}}{\mathrm{2}}\right]} \:\:+\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} −\:{H}_{\mathrm{2}\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]+\mathrm{1}} \\ $$$$ \\ $$
Answered by arcana last updated on 17/Oct/18
1)como u_n =Σ_(k=1) ^n (((−1)^k )/k) si n=2t es par (para algun t entero positivo) u_n =Σ_(k=1) ^(n/2) (((−1)^(2k−1) )/(2k−1))+Σ_(k=1) ^(n/2) (((−1)^(2k) )/(2k)) =−Σ_(k=1) ^(n/2) (1/(2k−1)) + Σ_(k=1) ^(n/2) (1/(2k)) =−Σ_(k=1) ^(n/2) (1/(2k−1)) + (1/2)H_(n/2) tomando m=2k−1, luego =−Σ_(m=1) ^(n−1) (1/m) + (1/2)H_(n/2) =(1/2)H_(n/2) −H_(n−1) si n=2t−1 es impar, tenemos u_n =Σ_(k=1) ^((n+1)/2) (((−1)^(2k−1) )/(2k−1))+Σ_(k=1) ^((n−1)/2) (((−1)^(2k) )/(2k)) =−Σ_(k=1) ^((n+1)/2) (1/(2k−1)) + Σ_(k=1) ^((n−1)/2) (1/(2k)) =−Σ_(k=1) ^((n+1)/2) (1/(2k−1)) + (1/2)H_((n−1)/2) tomando m=2k−1 =−Σ_(m=1) ^n (1/m) + (1/2)H_((n−1)/2) =−H_n +(1/2)H_((n−1)/2)
$$\left.\mathrm{1}\right){como}\:\:{u}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}} \\ $$$${si}\:{n}=\mathrm{2}{t}\:{es}\:{par}\:\left({para}\:{algun}\:{t}\:{entero}\:{positivo}\right) \\ $$$$ \\ $$$${u}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}/\mathrm{2}} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{\mathrm{2}{k}−\mathrm{1}} }{\mathrm{2}{k}−\mathrm{1}}+\underset{{k}=\mathrm{1}} {\overset{{n}/\mathrm{2}} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{\mathrm{2}{k}} }{\mathrm{2}{k}} \\ $$$$=−\underset{{k}=\mathrm{1}} {\overset{{n}/\mathrm{2}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\:+\:\underset{{k}=\mathrm{1}} {\overset{{n}/\mathrm{2}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{2}{k}} \\ $$$$=−\underset{{k}=\mathrm{1}} {\overset{{n}/\mathrm{2}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}/\mathrm{2}} \\ $$$${tomando}\:{m}=\mathrm{2}{k}−\mathrm{1},\:{luego} \\ $$$$=−\underset{{m}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\:\frac{\mathrm{1}}{{m}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}/\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}/\mathrm{2}} −{H}_{{n}−\mathrm{1}} \\ $$$$ \\ $$$${si}\:{n}=\mathrm{2}{t}−\mathrm{1}\:{es}\:{impar},\:{tenemos} \\ $$$${u}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{\left({n}+\mathrm{1}\right)/\mathrm{2}} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{\mathrm{2}{k}−\mathrm{1}} }{\mathrm{2}{k}−\mathrm{1}}+\underset{{k}=\mathrm{1}} {\overset{\left({n}−\mathrm{1}\right)/\mathrm{2}} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{\mathrm{2}{k}} }{\mathrm{2}{k}} \\ $$$$=−\underset{{k}=\mathrm{1}} {\overset{\left({n}+\mathrm{1}\right)/\mathrm{2}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\:+\:\underset{{k}=\mathrm{1}} {\overset{\left({n}−\mathrm{1}\right)/\mathrm{2}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{2}{k}} \\ $$$$=−\underset{{k}=\mathrm{1}} {\overset{\left({n}+\mathrm{1}\right)/\mathrm{2}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{H}_{\left({n}−\mathrm{1}\right)/\mathrm{2}} \\ $$$$ \\ $$$${tomando}\:{m}=\mathrm{2}{k}−\mathrm{1} \\ $$$$=−\underset{{m}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{{m}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}{H}_{\left({n}−\mathrm{1}\right)/\mathrm{2}} =−{H}_{{n}} +\frac{\mathrm{1}}{\mathrm{2}}{H}_{\left({n}−\mathrm{1}\right)/\mathrm{2}} \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 17/Oct/18
thank you Arkana for this work...
$${thank}\:{you}\:{Arkana}\:{for}\:{this}\:{work}… \\ $$
Answered by arcana last updated on 17/Oct/18
3. como (1/(1+x))=Σ_(k=0) ^∞ (−x)^k para ∣x∣<1 integrando a ambos lados vemos que ⇒ln (1+x)=Σ_(k=0) ^∞ (((−1)^(k+1) )/(k+1))∙x^(k+1) con −1<x≤1. Si x=1, entonces ln(2)=Σ_(k=0) ^∞ (((−1)^(k+1) )/(k+1))=Σu_n
$$\mathrm{3}.\:{como}\:\:\frac{\mathrm{1}}{\mathrm{1}+{x}}=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−{x}\right)^{{k}} \:\:{para}\:\mid{x}\mid<\mathrm{1} \\ $$$${integrando}\:{a}\:{ambos}\:{lados}\:{vemos}\:{que} \\ $$$$\Rightarrow{ln}\:\left(\mathrm{1}+{x}\right)=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} }{{k}+\mathrm{1}}\centerdot{x}^{{k}+\mathrm{1}} \\ $$$${con}\:−\mathrm{1}<{x}\leqslant\mathrm{1}.\:{Si}\:{x}=\mathrm{1},\:{entonces} \\ $$$$ \\ $$$${ln}\left(\mathrm{2}\right)=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} }{{k}+\mathrm{1}}=\Sigma{u}_{{n}} \\ $$

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