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Question-200976




Question Number 200976 by Blackpanther last updated on 27/Nov/23
Answered by Mathspace last updated on 28/Nov/23
Σ_(n=1) ^∞ ((cos(nπ))/(ln3))=Σ_(n=1) ^∞ (((−1)^n )/(ln3))  lim_(n→∞) (−1)^n ≠0 ⇒this serie  is divergente  2)Σ_(n=1) ^∞ n^(−(1/4)) =Σ_(n=1) ^∞ (1/n^(1/4) )  n^(1/4) <n^(1/2) ⇒(1/n^(1/4) )>(1/n^(1/2) )=(1/( (√n)))  but Σ(1/( (√n)))is div. ⇒Σn^(−(1/4))   is divergente  3)Σ(1/(nlogn))  let f(t)=(1/(tlogt))  (t≥2)  f^′ (t)=−((logt+1)/((tlogt)^2 ))<0 ⇒f is  decreasing so Σ u_n  and  ∫_2 ^∞ (dt/(tlogt)) have same nature  logt=u ⇒∫_2 ^∞ (dt/(tlogt))  =∫_(log2) ^∞ ((e^u du)/(e^u u))=∫_(log2) ^∞ (du/u)  =[logu]_(log2) ^∞ =+∞ ⇒  this serie is div.
$$\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{cos}\left({n}\pi\right)}{{ln}\mathrm{3}}=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{ln}\mathrm{3}} \\ $$$${lim}_{{n}\rightarrow\infty} \left(−\mathrm{1}\right)^{{n}} \neq\mathrm{0}\:\Rightarrow{this}\:{serie} \\ $$$${is}\:{divergente} \\ $$$$\left.\mathrm{2}\right)\sum_{{n}=\mathrm{1}} ^{\infty} {n}^{−\frac{\mathrm{1}}{\mathrm{4}}} =\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{n}^{\frac{\mathrm{1}}{\mathrm{4}}} } \\ $$$${n}^{\frac{\mathrm{1}}{\mathrm{4}}} <{n}^{\frac{\mathrm{1}}{\mathrm{2}}} \Rightarrow\frac{\mathrm{1}}{{n}^{\frac{\mathrm{1}}{\mathrm{4}}} }>\frac{\mathrm{1}}{{n}^{\frac{\mathrm{1}}{\mathrm{2}}} }=\frac{\mathrm{1}}{\:\sqrt{{n}}} \\ $$$${but}\:\Sigma\frac{\mathrm{1}}{\:\sqrt{{n}}}{is}\:{div}.\:\Rightarrow\Sigma{n}^{−\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$${is}\:{divergente} \\ $$$$\left.\mathrm{3}\right)\Sigma\frac{\mathrm{1}}{{nlogn}} \\ $$$${let}\:{f}\left({t}\right)=\frac{\mathrm{1}}{{tlogt}}\:\:\left({t}\geqslant\mathrm{2}\right) \\ $$$${f}^{'} \left({t}\right)=−\frac{{logt}+\mathrm{1}}{\left({tlogt}\right)^{\mathrm{2}} }<\mathrm{0}\:\Rightarrow{f}\:{is} \\ $$$${decreasing}\:{so}\:\Sigma\:{u}_{{n}} \:{and} \\ $$$$\int_{\mathrm{2}} ^{\infty} \frac{{dt}}{{tlogt}}\:{have}\:{same}\:{nature} \\ $$$${logt}={u}\:\Rightarrow\int_{\mathrm{2}} ^{\infty} \frac{{dt}}{{tlogt}} \\ $$$$=\int_{{log}\mathrm{2}} ^{\infty} \frac{{e}^{{u}} {du}}{{e}^{{u}} {u}}=\int_{{log}\mathrm{2}} ^{\infty} \frac{{du}}{{u}} \\ $$$$=\left[{logu}\right]_{{log}\mathrm{2}} ^{\infty} =+\infty\:\Rightarrow \\ $$$${this}\:{serie}\:{is}\:{div}. \\ $$$$ \\ $$
Answered by Mathspace last updated on 28/Nov/23
4)Σ(−1)^n (((2n)!)/((n!)^2 ))  u_n =(−1)^n (((2n)!)/((n!)^2 ))  ∣(u_(n+1) /u_n )∣=(((2n+2)!)/((n+1)!)^2 ))×((n!^2 )/((2n)!))  =(((2n+2)(2n+1))/((n+1)^2 ))→4>1  ⇒this serie is divergente
$$\left.\mathrm{4}\right)\Sigma\left(−\mathrm{1}\right)^{{n}} \frac{\left(\mathrm{2}{n}\right)!}{\left({n}!\right)^{\mathrm{2}} } \\ $$$${u}_{{n}} =\left(−\mathrm{1}\right)^{{n}} \frac{\left(\mathrm{2}{n}\right)!}{\left({n}!\right)^{\mathrm{2}} } \\ $$$$\mid\frac{{u}_{{n}+\mathrm{1}} }{{u}_{{n}} }\mid=\frac{\left(\mathrm{2}{n}+\mathrm{2}\right)!}{\left.\left({n}+\mathrm{1}\right)!\right)^{\mathrm{2}} }×\frac{{n}!^{\mathrm{2}} }{\left(\mathrm{2}{n}\right)!} \\ $$$$=\frac{\left(\mathrm{2}{n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\rightarrow\mathrm{4}>\mathrm{1} \\ $$$$\Rightarrow{this}\:{serie}\:{is}\:{divergente} \\ $$
Answered by Mathspace last updated on 28/Nov/23
5)Σ_(n=1) ^∞ ((6n+6)/((−1)^n ))=6Σ_(n=1) ^∞ (n+1)(−1)^n (n+1=p)  =6Σ_(p=2) ^∞ p(−1)^(p−1)   =lim_(n→+∞) 6Σ_(p=2) ^n p(−1)^(p−1)   we have Σ_(p=2) ^n  x^p   =((1−x^(n+1) )/(1−x))−1−x  =((1−x^(n+1) −(1−x)(1+x))/(1−x))  =((1−x^(n+1) −1+x^2 )/(1−x))=((x^2 −x^(n+1) )/(1−x))  and Σ_(p=2) ^n px^(p−1) =(d/dx)(((x^2 −x^(n+1) )/(1−x)))  =(((2x−(n+1)x^n )(1−x)+x^2 −x^(n+1) )/((1−x)^2 ))  =((2x−2x^2 −(n+1)x^n +(n+1)x^(n+1) +x^2 −x^(n+1) )/((1−x)^2 ))  =((2x−x^2 −nx^n +(n+1)x^(n+1) )/((1−x)^2 ))  Σ_2 ^n p(−1)^(p−1)   =((−2−1−n(−1)^n +(n+1)(−1)^(n+1) )/4)  →∞ ⇒this serie is divergente
$$\left.\mathrm{5}\right)\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{6}{n}+\mathrm{6}}{\left(−\mathrm{1}\right)^{{n}} }=\mathrm{6}\sum_{{n}=\mathrm{1}} ^{\infty} \left({n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}} \left({n}+\mathrm{1}={p}\right) \\ $$$$=\mathrm{6}\sum_{{p}=\mathrm{2}} ^{\infty} {p}\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} \\ $$$$={lim}_{{n}\rightarrow+\infty} \mathrm{6}\sum_{{p}=\mathrm{2}} ^{{n}} {p}\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} \\ $$$${we}\:{have}\:\sum_{{p}=\mathrm{2}} ^{{n}} \:{x}^{{p}} \\ $$$$=\frac{\mathrm{1}−{x}^{{n}+\mathrm{1}} }{\mathrm{1}−{x}}−\mathrm{1}−{x} \\ $$$$=\frac{\mathrm{1}−{x}^{{n}+\mathrm{1}} −\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}{\mathrm{1}−{x}} \\ $$$$=\frac{\mathrm{1}−{x}^{{n}+\mathrm{1}} −\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{1}−{x}}=\frac{{x}^{\mathrm{2}} −{x}^{{n}+\mathrm{1}} }{\mathrm{1}−{x}} \\ $$$${and}\:\sum_{{p}=\mathrm{2}} ^{{n}} {px}^{{p}−\mathrm{1}} =\frac{{d}}{{dx}}\left(\frac{{x}^{\mathrm{2}} −{x}^{{n}+\mathrm{1}} }{\mathrm{1}−{x}}\right) \\ $$$$=\frac{\left(\mathrm{2}{x}−\left({n}+\mathrm{1}\right){x}^{{n}} \right)\left(\mathrm{1}−{x}\right)+{x}^{\mathrm{2}} −{x}^{{n}+\mathrm{1}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}{x}−\mathrm{2}{x}^{\mathrm{2}} −\left({n}+\mathrm{1}\right){x}^{{n}} +\left({n}+\mathrm{1}\right){x}^{{n}+\mathrm{1}} +{x}^{\mathrm{2}} −{x}^{{n}+\mathrm{1}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}{x}−{x}^{\mathrm{2}} −{nx}^{{n}} +\left({n}+\mathrm{1}\right){x}^{{n}+\mathrm{1}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$$\sum_{\mathrm{2}} ^{{n}} {p}\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} \\ $$$$=\frac{−\mathrm{2}−\mathrm{1}−{n}\left(−\mathrm{1}\right)^{{n}} +\left({n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{4}} \\ $$$$\rightarrow\infty\:\Rightarrow{this}\:{serie}\:{is}\:{divergente} \\ $$$$ \\ $$
Answered by Mathspace last updated on 28/Nov/23
6) ((lnn)/n) is decreasing  ∫_2 ^∞ ((logx)/x)=_ [log^2 x]_2 ^∞ −∫_2 ^∞ ((logx)/x)dx  ⇒2∫_2 ^∞ ((logx)/x)dx=(1/2)[log^2 x]_2 ^∞ =∞  ⇒Σ((logn)/n) is div.
$$\left.\mathrm{6}\right)\:\frac{{lnn}}{{n}}\:{is}\:{decreasing} \\ $$$$\int_{\mathrm{2}} ^{\infty} \frac{{logx}}{{x}}=_{} \left[{log}^{\mathrm{2}} {x}\right]_{\mathrm{2}} ^{\infty} −\int_{\mathrm{2}} ^{\infty} \frac{{logx}}{{x}}{dx} \\ $$$$\Rightarrow\mathrm{2}\int_{\mathrm{2}} ^{\infty} \frac{{logx}}{{x}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\left[{log}^{\mathrm{2}} {x}\right]_{\mathrm{2}} ^{\infty} =\infty \\ $$$$\Rightarrow\Sigma\frac{{logn}}{{n}}\:{is}\:{div}. \\ $$$$ \\ $$

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