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.](Q201020.png)
$$\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{cos}\left({n}\pi\right)}{{ln}\mathrm{3}}=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{ln}\mathrm{3}} \\ $$$${lim}_{{n}\rightarrow\infty} \left(−\mathrm{1}\right)^{{n}} \neq\mathrm{0}\:\Rightarrow{this}\:{serie} \\ $$$${is}\:{divergente} \\ $$$$\left.\mathrm{2}\right)\sum_{{n}=\mathrm{1}} ^{\infty} {n}^{−\frac{\mathrm{1}}{\mathrm{4}}} =\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{n}^{\frac{\mathrm{1}}{\mathrm{4}}} } \\ $$$${n}^{\frac{\mathrm{1}}{\mathrm{4}}} <{n}^{\frac{\mathrm{1}}{\mathrm{2}}} \Rightarrow\frac{\mathrm{1}}{{n}^{\frac{\mathrm{1}}{\mathrm{4}}} }>\frac{\mathrm{1}}{{n}^{\frac{\mathrm{1}}{\mathrm{2}}} }=\frac{\mathrm{1}}{\:\sqrt{{n}}} \\ $$$${but}\:\Sigma\frac{\mathrm{1}}{\:\sqrt{{n}}}{is}\:{div}.\:\Rightarrow\Sigma{n}^{−\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$${is}\:{divergente} \\ $$$$\left.\mathrm{3}\right)\Sigma\frac{\mathrm{1}}{{nlogn}} \\ $$$${let}\:{f}\left({t}\right)=\frac{\mathrm{1}}{{tlogt}}\:\:\left({t}\geqslant\mathrm{2}\right) \\ $$$${f}^{'} \left({t}\right)=−\frac{{logt}+\mathrm{1}}{\left({tlogt}\right)^{\mathrm{2}} }<\mathrm{0}\:\Rightarrow{f}\:{is} \\ $$$${decreasing}\:{so}\:\Sigma\:{u}_{{n}} \:{and} \\ $$$$\int_{\mathrm{2}} ^{\infty} \frac{{dt}}{{tlogt}}\:{have}\:{same}\:{nature} \\ $$$${logt}={u}\:\Rightarrow\int_{\mathrm{2}} ^{\infty} \frac{{dt}}{{tlogt}} \\ $$$$=\int_{{log}\mathrm{2}} ^{\infty} \frac{{e}^{{u}} {du}}{{e}^{{u}} {u}}=\int_{{log}\mathrm{2}} ^{\infty} \frac{{du}}{{u}} \\ $$$$=\left[{logu}\right]_{{log}\mathrm{2}} ^{\infty} =+\infty\:\Rightarrow \\ $$$${this}\:{serie}\:{is}\:{div}. \\ $$$$ \\ $$
Answered by Mathspace last updated on 28/Nov/23
$$\left.\mathrm{4}\right)\Sigma\left(−\mathrm{1}\right)^{{n}} \frac{\left(\mathrm{2}{n}\right)!}{\left({n}!\right)^{\mathrm{2}} } \\ $$$${u}_{{n}} =\left(−\mathrm{1}\right)^{{n}} \frac{\left(\mathrm{2}{n}\right)!}{\left({n}!\right)^{\mathrm{2}} } \\ $$$$\mid\frac{{u}_{{n}+\mathrm{1}} }{{u}_{{n}} }\mid=\frac{\left(\mathrm{2}{n}+\mathrm{2}\right)!}{\left.\left({n}+\mathrm{1}\right)!\right)^{\mathrm{2}} }×\frac{{n}!^{\mathrm{2}} }{\left(\mathrm{2}{n}\right)!} \\ $$$$=\frac{\left(\mathrm{2}{n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\rightarrow\mathrm{4}>\mathrm{1} \\ $$$$\Rightarrow{this}\:{serie}\:{is}\:{divergente} \\ $$
Answered by Mathspace last updated on 28/Nov/23
$$\left.\mathrm{5}\right)\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{6}{n}+\mathrm{6}}{\left(−\mathrm{1}\right)^{{n}} }=\mathrm{6}\sum_{{n}=\mathrm{1}} ^{\infty} \left({n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}} \left({n}+\mathrm{1}={p}\right) \\ $$$$=\mathrm{6}\sum_{{p}=\mathrm{2}} ^{\infty} {p}\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} \\ $$$$={lim}_{{n}\rightarrow+\infty} \mathrm{6}\sum_{{p}=\mathrm{2}} ^{{n}} {p}\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} \\ $$$${we}\:{have}\:\sum_{{p}=\mathrm{2}} ^{{n}} \:{x}^{{p}} \\ $$$$=\frac{\mathrm{1}−{x}^{{n}+\mathrm{1}} }{\mathrm{1}−{x}}−\mathrm{1}−{x} \\ $$$$=\frac{\mathrm{1}−{x}^{{n}+\mathrm{1}} −\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}{\mathrm{1}−{x}} \\ $$$$=\frac{\mathrm{1}−{x}^{{n}+\mathrm{1}} −\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{1}−{x}}=\frac{{x}^{\mathrm{2}} −{x}^{{n}+\mathrm{1}} }{\mathrm{1}−{x}} \\ $$$${and}\:\sum_{{p}=\mathrm{2}} ^{{n}} {px}^{{p}−\mathrm{1}} =\frac{{d}}{{dx}}\left(\frac{{x}^{\mathrm{2}} −{x}^{{n}+\mathrm{1}} }{\mathrm{1}−{x}}\right) \\ $$$$=\frac{\left(\mathrm{2}{x}−\left({n}+\mathrm{1}\right){x}^{{n}} \right)\left(\mathrm{1}−{x}\right)+{x}^{\mathrm{2}} −{x}^{{n}+\mathrm{1}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}{x}−\mathrm{2}{x}^{\mathrm{2}} −\left({n}+\mathrm{1}\right){x}^{{n}} +\left({n}+\mathrm{1}\right){x}^{{n}+\mathrm{1}} +{x}^{\mathrm{2}} −{x}^{{n}+\mathrm{1}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}{x}−{x}^{\mathrm{2}} −{nx}^{{n}} +\left({n}+\mathrm{1}\right){x}^{{n}+\mathrm{1}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$$\sum_{\mathrm{2}} ^{{n}} {p}\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} \\ $$$$=\frac{−\mathrm{2}−\mathrm{1}−{n}\left(−\mathrm{1}\right)^{{n}} +\left({n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{4}} \\ $$$$\rightarrow\infty\:\Rightarrow{this}\:{serie}\:{is}\:{divergente} \\ $$$$ \\ $$
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.](Q201025.png)
$$\left.\mathrm{6}\right)\:\frac{{lnn}}{{n}}\:{is}\:{decreasing} \\ $$$$\int_{\mathrm{2}} ^{\infty} \frac{{logx}}{{x}}=_{} \left[{log}^{\mathrm{2}} {x}\right]_{\mathrm{2}} ^{\infty} −\int_{\mathrm{2}} ^{\infty} \frac{{logx}}{{x}}{dx} \\ $$$$\Rightarrow\mathrm{2}\int_{\mathrm{2}} ^{\infty} \frac{{logx}}{{x}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\left[{log}^{\mathrm{2}} {x}\right]_{\mathrm{2}} ^{\infty} =\infty \\ $$$$\Rightarrow\Sigma\frac{{logn}}{{n}}\:{is}\:{div}. \\ $$$$ \\ $$