If 0 < x < 1 and y = 1 2 x 2 + 2 3 x 3 + 3 4 x 4 + … … , then the value of e 1 + y at x…

If $0<x<1$ and $y=\frac{1}{2}{x}^2+\frac{2}{3}{x}^3+\frac{3}{4}{x}^4+\dots \dots$

$y=\frac{1}{2}{x}^2+\frac{2}{3}{x}^3+\frac{3}{4}{x}^4+\dots \dots$

$y=\left(1-\frac{1}{2}\right)x^2+\left(1-\frac{1}{3}\right)x^3+\left(1-\frac{1}{4}\right)x^4+\cdots \cdots ·\cdots$
$=\left(x^2+x^3+x^4+\cdots \cdots \right)-\left(\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\cdots \cdots \cdots \right)$
$=\frac{x^2}{1-x}+x-\left(x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\cdots \cdots \cdots \right)$

$=\frac{x}{1-x}+\ell n(1-x)$
Put $x=\frac{1}{2}$

$y=1-\ell n2$

Then, $e^{1+y}=e^{1+1-\ell n2}$

$=\frac{1}{2}{e}^2$