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could anyone tell me how to calculate these sums?I am not finding any usual way to calculate them.

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1  
What does $3.5.7$ even mean? –  filmor Oct 4 '12 at 9:27
3  
@filmor: Those are products. I suspect that this book would represent a common approximation to $\pi$ as $3\cdot1416$. –  Brian M. Scott Oct 4 '12 at 10:35

4 Answers 4

up vote 8 down vote accepted

5.6:

$$\begin{align*} \frac{1}{2\cdot3}+\frac{1}{4\cdot5}+\frac{1}{6\cdot7} &=\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\cdots\\ &=\sum_{n=2}^{\infty}\frac{(-1)^{n}}{n}\\ &=\sum_{n=2}^{\infty}\int_{0}^{-1}x^{n-1}dx\\ &=\int_{0}^{-1}\sum_{n=2}^{\infty}x^{n-1}dx\\ &=\int_{0}^{-1}\frac{x}{1-x}dx\\ &=1-\ln2 \end{align*}$$

5.8:

$$\begin{align*} \frac{1}{3}+\frac{1}{4}\cdot\frac{1}{2!}+\frac{1}{5}\cdot\frac{1}{3!}+\cdots &=\sum_{n=1}^{\infty}\frac{1}{n+2}\cdot\frac{1}{n!}\\ &=\sum_{n=1}^{\infty}\frac{n+1}{(n+2)!}\\ &=\sum_{n=1}^{\infty}[\frac{1}{(n+1)!}-\frac{1}{(n+2)!}]\\ &=\frac{1}{2} \end{align*}$$

The sum to $m$ terms is

$$\begin{align*} \sum_{n=1}^{m}\frac{1}{n+2}\cdot\frac{1}{n!} &=\sum_{n=1}^{m}\frac{n+1}{(n+2)!}\\ &=\sum_{n=1}^{m}[\frac{1}{(n+1)!}-\frac{1}{(n+2)!}]\\ &=\frac{1}{2}-\frac{1}{(m+2)!} \end{align*}$$

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$5.7$ is not clear to me still :( –  El Angel Exterminador Oct 4 '12 at 12:06
    
@Flute: Ok, 5.7 seems not very easy to work out as fast as the other two questions. I'll work on it and work it out as soon as possible. –  Alfred Chern Oct 4 '12 at 12:17
    
@Flute: I know how to deal with it now. Please wait. –  Alfred Chern Oct 4 '12 at 12:28
    
I myself has got it $(1+x)^{1/2}$ where $x=1$ is it that? –  El Angel Exterminador Oct 4 '12 at 13:31
    
@Flute: I think it not. How you got it? May be I can check it for you. –  Alfred Chern Oct 4 '12 at 14:00

5.7:

$$\begin{align*} 1+\frac{3}{4}+\frac{3\cdot5}{4\cdot8}+\cdots &=1+\sum_{n=1}^{\infty}\frac{3\cdot5\cdots(2n+1)}{4\cdot8\cdots4n}\\ &=1+\sum_{n=1}^{\infty}\frac{3\cdot5\cdots(2n+1)}{2\cdot4\cdots2n}(\frac{\sqrt{2}}{2})^{2n} \end{align*}$$

Set $f(x)=\sum_{n=1}^{\infty}\frac{3\cdot5\cdots(2n-1)}{2\cdot4\cdots2n}x^{2n+1}$, then $1+\frac{3}{4}+\frac{3\cdot5}{4\cdot8}+\cdots=1+f^{\prime}(\frac{\sqrt{2}}{2})$.

$$\begin{align*} f^{\prime}(x)&=\sum_{n=1}^{\infty}\frac{3\cdot5\cdots(2n+1)}{2\cdot4\cdots2n}x^{2n}\\ &=\sum_{n=1}^{\infty}\frac{3\cdot5\cdots(2n-1)\cdot(2n+1)}{2\cdot4\cdots2n}x^{2n}\\ &=\sum_{n=1}^{\infty}\frac{3\cdot5\cdots(2n-1)\cdot2n}{2\cdot4\cdots(2n-2)\cdot2n}x^{2n}+\sum_{n=1}^{\infty}\frac{3\cdot5\cdots(2n-1)}{2\cdot4\cdots2n}x^{2n}\\ &=x^{2}(1+\sum_{n=2}^{\infty}\frac{3\cdot5\cdots(2n-1)}{2\cdot4\cdots(2n-2)}x^{2n-2})+\frac{1}{x}f(x)\\ &=x^{2}(1+f^{\prime}(x))+\frac{1}{x}f(x) \end{align*}$$

Set $g(x)=f(x)+x$, then $g^{\prime}(x)=x^{2}g^{\prime}(x)+\frac{1}{x}g(x)$, by calculation: $$g^{\prime}(x)=\frac{1}{x(1-x^{2})}g(x)$$

$$(\frac{\sqrt{1-x^{2}}}{x}g(x))^{\prime}=0$$

$$\frac{\sqrt{1-x^{2}}}{x}g(x)=c$$

$$g(x)=c\frac{x}{\sqrt{1-x^{2}}}$$

$$g^{\prime}(x)=c\frac{1}{\sqrt{1-x^{2}}^{\frac{3}{2}}}$$

as $g^{\prime}(0)=f^{\prime}(0)+1=1$, so $c=1$, $1+\frac{3}{4}+\frac{3\cdot5}{4\cdot8}+\cdots=1+f^{\prime}(\frac{\sqrt{2}}{2})=g^{\prime}(\frac{\sqrt{2}}{2})=2\sqrt{2}$.

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This exercises seems to practise taylor expansions. For the last problem $$e^x = \sum _ { n = 1}^{\infty} \frac{x^n}{ n\mathrm{!}}$$ $$xe^x = \sum _ { n = 1}^{\infty} \frac{x^{n+1}}{ n\mathrm{!}}$$ now you can integrate and take the value at $1$

For the second you look at taylor expansion of $\arccos$ $$ \arccos(x) =\frac{\pi}{2}- \sum _ { n = 1}^{\infty} \frac{1}{4^n} \frac{2n \mathrm{!}}{n\mathrm{!}^2}\frac{1}{2n+1}x^{2n+1}$$ The general term of the second sum is $$ \frac{1 \cdot 3 \dots \cdot (2n+1)}{4^nn\mathrm{!}}=\frac{1 \cdot 3 \dots \cdot (2n)}{4^nn\mathrm{!}}\frac{1}{2^n n\mathrm{!}}=\frac{1}{2^{3n}}\frac{2n\mathrm{!}}{n\mathrm{!} ^2}$$ Now it is not very hard from the taylor expansion to get there.

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The denominator of the general term of 5.7 is $4^nn!$, not $4n!$. –  Brian M. Scott Oct 4 '12 at 11:00
    
Thank you, edited! –  clark Oct 4 '12 at 11:07

5.6.: Using $\displaystyle\frac1{n\cdot (n+1)} = \frac1n-\frac1{n+1}$: $$\frac12-\frac13+\frac14-\frac15\pm\cdots = 1-\log 2$$ Since $\log(1-x)= -\displaystyle\sum_{n\ge 1}\frac{x^n}n$, convergent at $x=-1$.

5.7.: Perhaps binomial series and generalized binomial coefficients help: $$\begin{pmatrix} -3/2\\n \end{pmatrix} = (-1)^n\cdot \frac{\overbrace{3/2\cdot 5/2\cdot 7/2\cdot..}^n}{n!}$$

5.8.: Observe that $\displaystyle\frac1{n+2}\cdot\frac1{n!} = \frac{n+1}{(n+2)!}$, and try to find a suitable power series..

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