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

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