Question

Given that x > 0, the sum $\sum _{n=1}^{\infty }{\left(\frac{x}{x+1}\right)}^{n-1}$ equals
(a) x

(b) x + 1

(c) $\frac{x}{2x+1}$

(d) $\frac{x+1}{2x+1}$

Answer 1

(b) x + 1

$$\begin{gathered} \sum _{n=1}^{\infty }{\left(\frac{x}{x+1}\right)}^{\left(n-1\right)}=1+\left(\frac{x}{x+1}\right)+{\left(\frac{x}{x+1}\right)}^2+{\left(\frac{x}{x+1}\right)}^3+{\left(\frac{x}{x+1}\right)}^4+...\infty \\ =\frac{1}{1-\left(\frac{x}{x+1}\right)}\left[\because \mathrm{it}\mathrm{is}\mathrm{a}\mathrm{G}.\mathrm{P}.\mathrm{with}a\mathit{}=1\mathrm{and}\mathrm{r}=\left(\frac{\mathrm{x}}{\mathrm{x}+1}\right)\right] \\ =\frac{\left(x+1\right)}{\left(x+1-x\right)} \\ =\frac{\left(x+1\right)}{1}=\left(x+1\right) \end{gathered}$$