Let-f-x-x-1-x-2-and-r-0-f-x-r-be-a-convergent-series-Find-the-value-of-x-such-that-n-0-r-0-x-1-x-2-r-n-4-

Question Number 122036 by physicstutes last updated on 13/Nov/20

Let f (x) = (\frac{x + 1}{x + 2}) and \underset{r = 0}{\sum^{\infty}} {[f (x)]}^{r} be a convergent series

Find the value of x such that

\underset{n = 0}{\sum^{\infty}} {[\underset{r = 0}{\sum^{\infty}} {(\frac{x + 1}{x + 2})}^{r}]}^{n} = 4

Answered by Dwaipayan Shikari last updated on 13/Nov/20

f (x) = \frac{x + 1}{x + 2}

\underset{r = 0}{\sum^{\infty}} {[f (x)]}^{r} = 1 + \frac{x + 1}{x + 2} + \dots = \frac{1}{1 - \frac{x + 1}{x + 2}} = x + 2

\underset{n = 0}{\sum^{\infty}} {(x + 2)}^{n} = \frac{1}{1 - (x + 2)} = \frac{1}{- (x + 1)}

\frac{1}{- (x + 1)} = 4 \Rightarrow - (x + 1) = \frac{1}{4} \Rightarrow x = - \frac{5}{4}

Commented by physicstutes last updated on 13/Nov/20

Fantastic result . What i execpted . Thanks alot .