n-1-1-n-1-n-n-2-

Question Number 161623 by amin96 last updated on 20/Dec/21

$$\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}+\mathrm{1}} }{\boldsymbol{{n}}\left(\boldsymbol{{n}}+\mathrm{2}\right)}=? \\ $$

Answered by TheSupreme last updated on 20/Dec/21

(A/n)+(B/(n+2))=(((−1)^(n+1) )/(n(n+2))) An+2A+Bn=(−1)^(n+1) A=−B=(((−1)^(n+1) )/2) Σ_(n=1) ^∞ (((−1)^(n+1) )/2)(1/n)−(((−1)^(n+1) )/2)(1/(n+2)) s_n =(1/2)−(1/3)+(((−1)^n )/(n+2)) lim s_n =(1/2)−(1/3)=(1/6)

$$\frac{{A}}{{n}}+\frac{{B}}{{n}+\mathrm{2}}=\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}\left({n}+\mathrm{2}\right)} \\ $$$${An}+\mathrm{2}{A}+{Bn}=\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \\ $$$${A}=−{B}=\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{2}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{2}}\frac{\mathrm{1}}{{n}}−\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{2}}\frac{\mathrm{1}}{{n}+\mathrm{2}} \\ $$$${s}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{2}} \\ $$$${lim}\:{s}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$

Commented by mr W last updated on 21/Dec/21

how did you come from An+2A+Bn=(−1)^(n+1) to A=−B=(((−1)^(n+1) )/2) ?

$${how}\:{did}\:{you}\:{come}\:{from} \\ $$$${An}+\mathrm{2}{A}+{Bn}=\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \\ $$$${to}\: \\ $$$${A}=−{B}=\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{2}}\:? \\ $$

Answered by mathmax by abdo last updated on 20/Dec/21

S_n =Σ_(k=1) ^n (((−1)^(k+1) )/(k(k+2))) ⇒S_n =(1/2)Σ_(k=1) ^n ((1/k)−(1/(k+2)))(−1)^(k+1) =−(1/2)Σ_(k=1) ^n (((−1)^k )/k) +(1/2)Σ_(k=1) ^(n ) (((−1)^k )/(k+2)) (→k+2=p) =−(1/2)Σ_(k=1) ^n (((−1)^k )/k)+(1/2)Σ_(p=3) ^(n+2) (((−1)^p )/p) =−(1/2)Σ_(k=1) ^n (((−1)^k )/k)+(1/2){Σ_(p=1) ^n (((−1)^p )/p)−(−1+(1/2))+(((−1)^n )/(n+2))} =(1/2){(1/2)+(((−1)^n )/(n+2))} ⇒lim_(n→+∞) S_n =(1/4)

$$\mathrm{S}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}+\mathrm{1}} }{\mathrm{k}\left(\mathrm{k}+\mathrm{2}\right)}\:\Rightarrow\mathrm{S}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{2}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left(\frac{\mathrm{1}}{\mathrm{k}}−\frac{\mathrm{1}}{\mathrm{k}+\mathrm{2}}\right)\left(−\mathrm{1}\right)^{\mathrm{k}+\mathrm{1}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{k}}\:+\frac{\mathrm{1}}{\mathrm{2}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}\:} \frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{k}+\mathrm{2}}\:\left(\rightarrow\mathrm{k}+\mathrm{2}=\mathrm{p}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{k}}+\frac{\mathrm{1}}{\mathrm{2}}\sum_{\mathrm{p}=\mathrm{3}} ^{\mathrm{n}+\mathrm{2}} \frac{\left(−\mathrm{1}\right)^{\mathrm{p}} }{\mathrm{p}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{k}}+\frac{\mathrm{1}}{\mathrm{2}}\left\{\sum_{\mathrm{p}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{p}} }{\mathrm{p}}−\left(−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}+\mathrm{2}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}+\mathrm{2}}\right\}\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{S}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$

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