sum of the series 
could anyone tell me how to calculate these sums?I am not finding any usual way to calculate them.
 A: 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
$$\begin{align*}
\sum_{n=1}^{m}\frac{1}{n+2}\cdot\frac{1}{n!}
&=\sum_{n=1}^{m}\frac{n+1}{(n+2)!}\\
&=\sum_{n=1}^{m}[\frac{1}{(n+1)!}-\frac{1}{(n+2)!}]\\
&=\frac{1}{2}-\frac{1}{(m+2)!}
\end{align*}$$
A: 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..
A: 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.
A: 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 $c=1$, $1+\frac{3}{4}+\frac{3\cdot5}{4\cdot8}+\cdots=1+f^{\prime}(\frac{\sqrt{2}}{2})=g^{\prime}(\frac{\sqrt{2}}{2})=2\sqrt{2}$.
