Alg9.3.1

From Exampleproblems

Find the sum of the infinite series $\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...\,$

Rearranging the each fraction as a composition of two fractions, it looks like

$\left[\frac{1}{1}-\frac{1}{2}\right]-\left[\frac{1}{2}-\frac{1}{3}\right]+\left[\frac{1}{3}-\frac{1}{4}\right]+....... \,$

Now rearranging $1-\frac{1}{2}-\frac{1}{2}+\frac{1}{3}+\frac{1}{3}-\frac{1}{4}-\frac{1}{4}+...\,$

$1-(2)\cdot\frac{1}{2}+(2)\cdot\frac{1}{3}-(2)\cdot\frac{1}{4}+...\,$

$(2)(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...)-1\,$

Using the relation $\ln(1+x)=x-x^2\frac{1}{2}+x^3\frac{1}{3}-x^4\frac{1}{4}+..., -1<x\le 1\,$ the above equation can be written $2\ln(1+1)-\ln e\,$

Further simplifying $\ln 4-\ln e\,$

Solution is $\ln\frac{4}{e}\,$