Menu Close

let-w-n-k-2-n-1-k-2-1-find-lim-n-w-n-

Tinku Tara 0 Comments
Pin




Question Number 30524 by abdo imad last updated on 22/Feb/18
let w_n = Σ_(k=2) ^n (1/(k^2 −1)) find lim_(n→∞) w_n .
$${let}\:{w}_{{n}} =\:\sum_{{k}=\mathrm{2}} ^{{n}} \:\:\:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} −\mathrm{1}}\:\:{find}\:{lim}_{{n}\rightarrow\infty} \:{w}_{{n}} \:. \\ $$
Commented by abdo imad last updated on 23/Feb/18
the serie w_n is positif how to find a negatif value sir?...
$${the}\:{serie}\:{w}_{{n}} {is}\:{positif}\:{how}\:{to}\:{find}\:{a}\:{negatif}\:{value}\:{sir}?… \\ $$
Commented by abdo imad last updated on 24/Feb/18
we have w_n = Σ_(k=2) ^n (1/(k^2 −1))=(1/2)Σ_(k=2) ^n ((1/(k−1)) −(1/(k+1))) = (1/2) Σ_(k=2) ^n (1/(k−1)) −(1/2)Σ_(k=2) ^n (1/(k+1)) =(1/2)Σ_(k=1) ^(n−1) (1/k) −(1/2) Σ_(k=3) ^(n+1) (1/k) but we have Σ_(k=1) ^(n−1) (1/k)=H_(n−1) and Σ_(k=3) ^(n+1) (1/k)=H_(n+1) −(3/2) ⇒ w_n =(1/2)H_(n−1) −(1/2) H_(n+1) +(3/4)=(1/2)(H_(n−1) −H_(n+1) ) +(3/4) H_(n−1) =ln(n−1) +γ +o((1/n)) H_(n+1) =ln(n+1) +γ +o((1/n)) ⇒H_(n−1) −H_(n+1) =ln(((n−1)/(n+1))) +o((1/n)) ⇒lim_(n→∞) H_(n−1) −H_(n+1) =0 ⇒lim_(n→∞) w_n =(3/4) .
$${we}\:{have}\:{w}_{{n}} =\:\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} −\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{2}} ^{{n}} \left(\frac{\mathrm{1}}{{k}−\mathrm{1}}\:−\frac{\mathrm{1}}{{k}+\mathrm{1}}\right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}−\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{\mathrm{1}}{{k}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{3}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{{k}}\:{but}\:{we}\:{have} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\frac{\mathrm{1}}{{k}}={H}_{{n}−\mathrm{1}} \:{and}\:\sum_{{k}=\mathrm{3}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{{k}}={H}_{{n}+\mathrm{1}} \:−\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow \\ $$$${w}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}−\mathrm{1}} \:−\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}+\mathrm{1}} \:+\frac{\mathrm{3}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}}\left({H}_{{n}−\mathrm{1}} \:−{H}_{{n}+\mathrm{1}} \right)\:+\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${H}_{{n}−\mathrm{1}} ={ln}\left({n}−\mathrm{1}\right)\:+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right) \\ $$$${H}_{{n}+\mathrm{1}} \:={ln}\left({n}+\mathrm{1}\right)\:+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:\Rightarrow{H}_{{n}−\mathrm{1}} \:−{H}_{{n}+\mathrm{1}} ={ln}\left(\frac{{n}−\mathrm{1}}{{n}+\mathrm{1}}\right)\:+{o}\left(\frac{\mathrm{1}}{{n}}\right) \\ $$$$\Rightarrow{lim}_{{n}\rightarrow\infty} \:{H}_{{n}−\mathrm{1}} \:−{H}_{{n}+\mathrm{1}} =\mathrm{0}\:\Rightarrow{lim}_{{n}\rightarrow\infty} {w}_{{n}} =\frac{\mathrm{3}}{\mathrm{4}}\:. \\ $$
Answered by sma3l2996 last updated on 23/Feb/18
w_n =Σ_(k=2) ^n (1/(k^2 −1))=Σ_(k=2) ^n (1/((k−1)(k+1))) we have (1/((k+1)(k−1)))=(1/2)((1/(k−1))−(1/(k+1))) so w_n =(1/2)(Σ_(k=2) ^n (1/(k−1))−Σ_(k=2) ^n (1/(k+1))) i=k−1 ; j=k+1 w_n =(1/2)(Σ_(i=1) ^(n−1) (1/i)−Σ_(j=3) ^(n+1) (1/j)) w_n =(1/2)(Σ_(i=1) ^(n−1) (1/i)−(Σ_(j=1) ^(n−1) (1/j)+(1/n)+(1/(n+1))−1−(1/2))) w_n =(1/2)(Σ_(i=1) ^(n−1) (1/i)−Σ_(j=1) ^(n−1) (1/j)+((2n+1)/(n(n+1)))−(3/2)) w_n =(1/2)(((2n+1)/(n(n+1)))−(3/2)) so lim_(n→∞) w_n =lim_(n→∞) ((2n)/(2n^2 ))−(3/4)=−(3/4)
$${w}_{{n}} =\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} −\mathrm{1}}=\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\left({k}−\mathrm{1}\right)\left({k}+\mathrm{1}\right)} \\ $$$${we}\:{have}\:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)\left({k}−\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{k}−\mathrm{1}}−\frac{\mathrm{1}}{{k}+\mathrm{1}}\right) \\ $$$${so}\:\:{w}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}−\mathrm{1}}−\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}+\mathrm{1}}\right) \\ $$$${i}={k}−\mathrm{1}\:;\:{j}={k}+\mathrm{1} \\ $$$${w}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\underset{{i}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{i}}−\underset{{j}=\mathrm{3}} {\overset{{n}+\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{j}}\right) \\ $$$${w}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\underset{{i}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{i}}−\left(\underset{{j}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{j}}+\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}+\mathrm{1}}−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$${w}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\underset{{i}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{i}}−\underset{{j}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{j}}+\frac{\mathrm{2}{n}+\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}−\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$${w}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}{n}+\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}−\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$${so}\:\:\underset{{n}\rightarrow\infty} {{lim}w}_{{n}} =\underset{{n}\rightarrow\infty} {{lim}}\frac{\mathrm{2}{n}}{\mathrm{2}{n}^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{4}}=−\frac{\mathrm{3}}{\mathrm{4}} \\ $$
Commented by sma3l2996 last updated on 24/Feb/18
w_n =(1/2)(Σ_(i=1) ^(n−1) (1/i)−(Σ_(j=1) ^(n−1) (1/j)+((2n+1)/(n(n+1)))−(3/2))) =(1/2)(Σ_(i=1) ^(n−1) (1/i)−Σ_(j=1) ^(n−1) −((2n+1)/(n(n+1)))+(3/2)) w_n =(1/2)((3/2)−((2n+1)/(n(n+1)))) so lim_(n→∞) w_n =(1/2)((3/2)−lim_(n→∞) ((2n)/n^2 ))=(3/4)
$${w}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\underset{{i}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{i}}−\left(\underset{{j}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{j}}+\frac{\mathrm{2}{n}+\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}−\frac{\mathrm{3}}{\mathrm{2}}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\underset{{i}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{i}}−\underset{{j}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}−\frac{\mathrm{2}{n}+\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}+\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$${w}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{2}{n}+\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}\right) \\ $$$${so}\:\underset{{n}\rightarrow\infty} {{lim}w}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{3}}{\mathrm{2}}−\underset{{n}\rightarrow\infty} {{lim}}\frac{\mathrm{2}{n}}{{n}^{\mathrm{2}} }\right)=\frac{\mathrm{3}}{\mathrm{4}} \\ $$
Commented by sma3l2996 last updated on 24/Feb/18
I did mistake on line 7
$${I}\:{did}\:{mistake}\:{on}\:{line}\:\mathrm{7} \\ $$
Commented by prof Abdo imad last updated on 24/Feb/18
nevermind sir...
$${nevermind}\:{sir}… \\ $$

Pin

Leave a Reply

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