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Question Number 178254 by lapache last updated on 14/Oct/22

Calcul 1−Σ_(n=1) ^(+∝) (2/(4n^2 −1)) 2−Σ_(n=1) ^(+∝) (n^2 /(n!)) 3− Σ_(n=1) ^(+∝) (n^3 /(n!)) 4− Σ_(n=1) ^(+∝) ln((n/(n−1)))

$${Calcul} \\ $$$$\mathrm{1}−\underset{{n}=\mathrm{1}} {\overset{+\propto} {\sum}}\frac{\mathrm{2}}{\mathrm{4}{n}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\mathrm{2}−\underset{{n}=\mathrm{1}} {\overset{+\propto} {\sum}}\frac{{n}^{\mathrm{2}} }{{n}!} \\ $$$$\mathrm{3}−\:\underset{{n}=\mathrm{1}} {\overset{+\propto} {\sum}}\frac{{n}^{\mathrm{3}} }{{n}!} \\ $$$$\mathrm{4}−\:\underset{{n}=\mathrm{1}} {\overset{+\propto} {\sum}}{ln}\left(\frac{{n}}{{n}−\mathrm{1}}\right) \\ $$

Answered by mr W last updated on 14/Oct/22

(2) e^x =Σ_(n=0) ^∞ (x^n /(n!)) (e^x )′=Σ_(n=0) ^∞ ((x^n /(n!)))′ e^x =Σ_(n=1) ^∞ ((nx^(n−1) )/(n!)) xe^x =Σ_(n=1) ^∞ ((nx^n )/(n!)) (xe^x )′=Σ_(n=1) ^∞ (((nx^n )/(n!)))′ (x+1)e^x =Σ_(n=1) ^∞ ((n^2 x^(n−1) )/(n!)) with x=1: ⇒2e=Σ_(n=1) ^∞ (n^2 /(n!)) (3) similarly

$$\left(\mathrm{2}\right) \\ $$$${e}^{{x}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}!} \\ $$$$\left({e}^{{x}} \right)'=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{{x}^{{n}} }{{n}!}\right)' \\ $$$${e}^{{x}} =\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{nx}^{{n}−\mathrm{1}} }{{n}!} \\ $$$${xe}^{{x}} =\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{nx}^{{n}} }{{n}!} \\ $$$$\left({xe}^{{x}} \right)'=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{{nx}^{{n}} }{{n}!}\right)' \\ $$$$\left({x}+\mathrm{1}\right){e}^{{x}} =\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{2}} {x}^{{n}−\mathrm{1}} }{{n}!} \\ $$$${with}\:{x}=\mathrm{1}: \\ $$$$\Rightarrow\mathrm{2}{e}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{2}} }{{n}!} \\ $$$$\left(\mathrm{3}\right)\:{similarly} \\ $$

Commented by Tawa11 last updated on 14/Oct/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by mr W last updated on 14/Oct/22

(4) =lim_(n→∞) (Σ_(k=2) ^n ln (k/(k−1))) =lim_(n→∞) (ln (2/1)×(3/2)×...×(n/(n−1))) =lim_(n→∞) (ln n) =∞

$$\left(\mathrm{4}\right) \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}\mathrm{ln}\:\frac{{k}}{{k}−\mathrm{1}}\right) \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{ln}\:\frac{\mathrm{2}}{\mathrm{1}}×\frac{\mathrm{3}}{\mathrm{2}}×...×\frac{{n}}{{n}−\mathrm{1}}\right) \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{ln}\:{n}\right) \\ $$$$=\infty \\ $$

Answered by mr W last updated on 14/Oct/22

(1) =Σ_(n=1) ^∞ ((1/(2n−1))−(1/(2n+1))) =((1/1)−(1/3))+((1/3)−(1/5))+((1/5)−(1/7))+... =1

$$\left(\mathrm{1}\right) \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right) \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{5}}\right)+\left(\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{7}}\right)+... \\ $$$$=\mathrm{1} \\ $$

Commented by lapache last updated on 14/Oct/22

Thank

$${Thank} \\ $$

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