Given that x > 0, the sum

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

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

 

Solution:

(b) x + 1

$\sum_{n=1}^{\infty}\left(\frac{x}{x+1}\right)^{(n-1)}=1+\left(\frac{x}{x+1}\right)+\left(\frac{x}{x+1}\right)^{2}+\left(\frac{x}{x+1}\right)^{3}+\left(\frac{x}{x+1}\right)^{4}+\ldots \infty$

$=\frac{1}{1-\left(\frac{x}{x+1}\right)} \quad\left[\because\right.$ it is a G.P. with $a=1$ and $\left.\mathrm{r}=\left(\frac{\mathrm{x}}{\mathrm{x}+1}\right)\right]$

$=\frac{(x+1)}{(x+1-x)}$

$=\frac{(x+1)}{1}=(x+1)$

 

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