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

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

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

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

Commented by Tawa11 last updated on 14/Oct/22

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

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

$$\left(4\right)$$$$=\underset{n\to \mathrm{\infty }}{\lim }\left(\underset{k=2}{\sum ^n}\ln \frac{k}{k-1}\right)$$$$=\underset{n\to \mathrm{\infty }}{\lim }(\ln \frac{2}{1}\times \frac{3}{2}\times ...\times \frac{n}{n-1})$$$$=\underset{n\to \mathrm{\infty }}{\lim }\left(\ln n\right)$$$$=\mathrm{\infty }$$

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

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

Commented by lapache last updated on 14/Oct/22

$$Thank$$

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