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Question Number 140833 by liberty last updated on 13/May/21
Determine if the series Σ_(n=1) ^∞ a_n defined by the formula converges or diverges . a_1 = 4 , a_(n+1) = ((10+sin n)/n). a_n
$$\mathrm{Determine}\:\mathrm{if}\:\mathrm{the}\:\mathrm{series}\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\mathrm{a}_{\mathrm{n}} \\ $$$$\mathrm{defined}\:\mathrm{by}\:\mathrm{the}\:\:\mathrm{formula}\:\mathrm{converges}\:\mathrm{or} \\ $$$$\mathrm{diverges}\:.\:\mathrm{a}_{\mathrm{1}} =\:\mathrm{4}\:,\:\mathrm{a}_{\mathrm{n}+\mathrm{1}} \:=\:\frac{\mathrm{10}+\mathrm{sin}\:\mathrm{n}}{\mathrm{n}}.\:\mathrm{a}_{\mathrm{n}} \\ $$
Commented by liberty last updated on 13/May/21
can anyone help me
can anyone help me
Answered by TheSupreme last updated on 13/May/21
(a_(n+1) /a_n )=((10+sin(n))/n) lim (a_(n+1) /a_n )=0 → series converges
$$\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }=\frac{\mathrm{10}+{sin}\left({n}\right)}{{n}} \\ $$$${lim}\:\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }=\mathrm{0}\:\rightarrow\:{series}\:{converges} \\ $$
Commented by liberty last updated on 13/May/21
why lim (a_(n+1) /a_n ) = 0? as n→?
$$\mathrm{why}\:\mathrm{lim}\:\frac{\mathrm{a}_{\mathrm{n}+\mathrm{1}} }{\mathrm{a}_{\mathrm{n}} }\:=\:\mathrm{0}?\:\mathrm{as}\:\mathrm{n}\rightarrow? \\ $$
Answered by mathmax by abdo last updated on 13/May/21
a_(n+1) =((10+sin(n))/n) a_n ⇒(a_(n+1) /a_n )=((10+sin(n))/n) ⇒ Π_(k=1) ^(n−1) (a_(k+1) /a_k ) =Π_(k=1) ^(n−1) (((10+sin(k))/k)) ⇒(a_2 /a_1 ).(a_3 /a_2 ).....(a_n /a_(n−1) )=(1/((n−1)!))Π_(k=1) ^(n−1) (10+sin(k)) ⇒ a_n =(4/((n−1)!))Π_(k=1) ^(n−1) (10+sin(k)) ⇒∣a_n ∣≤((4.11^n )/((n−1)!)) =v_n (v_(n+1) /v_n ) =((4.11^(n+1) )/(n!))×(((n−1)!)/(4.11^n )) =((11)/n)→0(n→∞) ⇒Σ v_n converges ⇒ Σ a_n cv.
$$\mathrm{a}_{\mathrm{n}+\mathrm{1}} =\frac{\mathrm{10}+\mathrm{sin}\left(\mathrm{n}\right)}{\mathrm{n}}\:\mathrm{a}_{\mathrm{n}} \:\Rightarrow\frac{\mathrm{a}_{\mathrm{n}+\mathrm{1}} }{\mathrm{a}_{\mathrm{n}} }=\frac{\mathrm{10}+\mathrm{sin}\left(\mathrm{n}\right)}{\mathrm{n}}\:\Rightarrow \\ $$$$\prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\mathrm{a}_{\mathrm{k}+\mathrm{1}} }{\mathrm{a}_{\mathrm{k}} }\:=\prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}−\mathrm{1}} \left(\frac{\mathrm{10}+\mathrm{sin}\left(\mathrm{k}\right)}{\mathrm{k}}\right)\:\Rightarrow\frac{\mathrm{a}_{\mathrm{2}} }{\mathrm{a}_{\mathrm{1}} }.\frac{\mathrm{a}_{\mathrm{3}} }{\mathrm{a}_{\mathrm{2}} }…..\frac{\mathrm{a}_{\mathrm{n}} }{\mathrm{a}_{\mathrm{n}−\mathrm{1}} }=\frac{\mathrm{1}}{\left(\mathrm{n}−\mathrm{1}\right)!}\prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}−\mathrm{1}} \left(\mathrm{10}+\mathrm{sin}\left(\mathrm{k}\right)\right)\:\Rightarrow \\ $$$$\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{4}}{\left(\mathrm{n}−\mathrm{1}\right)!}\prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}−\mathrm{1}} \left(\mathrm{10}+\mathrm{sin}\left(\mathrm{k}\right)\right)\:\Rightarrow\mid\mathrm{a}_{\mathrm{n}} \mid\leqslant\frac{\mathrm{4}.\mathrm{11}^{\mathrm{n}} }{\left(\mathrm{n}−\mathrm{1}\right)!}\:=\mathrm{v}_{\mathrm{n}} \\ $$$$\frac{\mathrm{v}_{\mathrm{n}+\mathrm{1}} }{\mathrm{v}_{\mathrm{n}} }\:=\frac{\mathrm{4}.\mathrm{11}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}!}×\frac{\left(\mathrm{n}−\mathrm{1}\right)!}{\mathrm{4}.\mathrm{11}^{\mathrm{n}} }\:=\frac{\mathrm{11}}{\mathrm{n}}\rightarrow\mathrm{0}\left(\mathrm{n}\rightarrow\infty\right)\:\Rightarrow\Sigma\:\mathrm{v}_{\mathrm{n}} \:\mathrm{converges}\:\Rightarrow \\ $$$$\Sigma\:\mathrm{a}_{\mathrm{n}} \:\mathrm{cv}. \\ $$
Answered by physicstutes last updated on 13/May/21
a_(n+1) = ((10 + sin n)/n) a_n ⇒ (a_(n+1) /a_n ) = ((10 + sin n)/n) = ((10)/(n )) + ((sin n)/n) lim_(n→∞) (a_(n+1) /a_n ) = lim_(n→∞) (((10)/n)+((sin n)/n)) = 0 + 0 = 0 < 1 ⇒ Σ_(n=1) ^∞ a_n converges by the ratio test.
$${a}_{{n}+\mathrm{1}} \:=\:\frac{\mathrm{10}\:+\:\mathrm{sin}\:{n}}{{n}}\:{a}_{{n}} \\ $$$$\Rightarrow\:\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }\:=\:\frac{\mathrm{10}\:+\:\mathrm{sin}\:{n}}{{n}}\:=\:\frac{\mathrm{10}}{{n}\:}\:+\:\frac{\mathrm{sin}\:{n}}{{n}} \\ $$$$\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{10}}{{n}}+\frac{\mathrm{sin}\:{n}}{{n}}\right)\:=\:\mathrm{0}\:+\:\mathrm{0}\:\:=\:\mathrm{0}\:<\:\mathrm{1} \\ $$$$\Rightarrow\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} \:\mathrm{converges}\:\mathrm{by}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{test}. \\ $$

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