If $$0 < x < 1$$ and $$y = \frac{1}{2}x^2 + \frac{2}{3}x^3 + \frac{3}{4}x^4 + \ldots$$, then the value of $$e^{1+y}$$ at $$x = \frac{1}{2}$$ is:

We are given the series

$$y \;=\; \frac12\,x^2 \;+\; \frac23\,x^3 \;+\; \frac34\,x^4 \;+\; \ldots$$

and we need the numerical value of $$e^{1+y}$$ when $$x=\dfrac12$$, keeping in mind that $$0<x<1$$ so every infinite series we write will certainly converge.

First we recognize the pattern of the coefficients. For the general $$k^{\text{th}}$$ term (starting with power $$x^2$$) the coefficient is

$$\frac{k-1}{k}$$

because when $$k=2,3,4,\ldots$$ we obtain respectively $$\dfrac12,\dfrac23,\dfrac34,\ldots$$ exactly as in the question. Hence we can rewrite $$y$$ compactly as

$$y \;=\; \sum_{k=2}^{\infty} \frac{k-1}{k}\,x^{\,k}.$$

Next we split the fraction into two simpler parts:

$$\frac{k-1}{k}\;=\;1-\frac1k.$$

Substituting this into the series gives

$$y \;=\; \sum_{k=2}^{\infty}\Bigl(1-\frac1k\Bigr)x^{\,k} \;=\;\sum_{k=2}^{\infty}x^{\,k}\;-\;\sum_{k=2}^{\infty}\frac{x^{\,k}}{k}.$$

Now we evaluate each sum separately.

We have a standard geometric‐series formula: for $$|x|<1$$,

$$\sum_{k=0}^{\infty}x^{\,k}\;=\;\frac1{1-x}.$$

So, removing the first two terms $$k=0,1$$ from that full series, we obtain

$$\sum_{k=2}^{\infty}x^{\,k} =\;\Bigl(\frac1{1-x}\Bigr)\;-\;1\;-\;x =\;\frac{1-(1-x)-x(1-x)}{1-x} =\;\frac{x^{2}}{1-x}.$$

(A quicker way is to notice directly that the first term present is $$x^{2}$$, so multiplying the usual remainder formula by $$x^{2}$$ also gives $$\dfrac{x^{2}}{1-x}$$.)

For the second sum we recall the Taylor expansion of the natural logarithm valid for $$|x|<1$$:

$$-\ln(1-x)\;=\;\sum_{k=1}^{\infty}\frac{x^{\,k}}{k}.$$

If we leave out the first term $$k=1$$ of that series, we have

$$\sum_{k=2}^{\infty}\frac{x^{\,k}}{k} =\;\Bigl(-\ln(1-x)\Bigr)\;-\;\frac{x^{1}}{1} =\;-\ln(1-x)\;-\;x.$$

Substituting both partial sums back into the expression for $$y$$ we get

$$\begin{aligned} y &=\;\frac{x^{2}}{1-x}\;-\;\Bigl(-\ln(1-x)\;-\;x\Bigr)\\[4pt] &=\;\frac{x^{2}}{1-x}\;+\;\ln(1-x)\;+\;x. \end{aligned}$$

Now we substitute the specific value $$x=\dfrac12$$. Step by step:

$$x\;=\;\frac12,\qquad 1-x\;=\;\frac12,\qquad x^{2}\;=\;\frac14.$$

Compute the first fraction:

$$\frac{x^{2}}{1-x} =\;\frac{\tfrac14}{\tfrac12} =\;\frac14\;\times\;2 =\;\frac12.$$

Compute the logarithmic term:

$$\ln(1-x)\;=\;\ln\!\Bigl(\tfrac12\Bigr) =\;-\ln 2.$$

The linear term is simply

$$x\;=\;\frac12.$$

Adding all three contributions:

$$\begin{aligned} y &=\;\frac12\;+\;(-\ln 2)\;+\;\frac12\\[4pt] &=\;1\;-\;\ln 2. \end{aligned}$$

Next we need $$e^{1+y}$$. First find $$1+y$$:

$$1+y\;=\;1\;+\;\bigl(1-\ln 2\bigr)\;=\;2\;-\;\ln 2.$$

Using the law of exponents $$e^{a-b}=e^{a}\,e^{-b}$$ we obtain

$$e^{1+y} =\;e^{\,2-\ln 2} =\;e^{\,2}\,e^{-\ln 2}.$$

The property $$e^{\ln k}=k$$ gives $$e^{-\ln 2}=2^{-1}=\dfrac12$$. Therefore

$$e^{1+y}\;=\;e^{2}\times\frac12\;=\;\frac12\,e^{2}.$$

Hence, the correct answer is Option A.