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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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    @filmor: Those are products. I suspect that this book would represent a common approximation to $\pi$ as $3\cdot1416$.2012-10-04

4 Answers 4

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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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    @Flute: I had give the solution in another answer, you can refer to it.2012-10-04
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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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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 $4n!$.2012-10-04
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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 $c=1$, $1+\frac{3}{4}+\frac{3\cdot5}{4\cdot8}+\cdots=1+f^{\prime}(\frac{\sqrt{2}}{2})=g^{\prime}(\frac{\sqrt{2}}{2})=2\sqrt{2}$.