Exponentiated Geometric Sum

Let's start from the well known expansion

$\frac{1}{1-x}=1+x^2+x^3+\dots\text{ provided }|{x}|<1\dots(1)$

Differentiating both sides shows that

$\frac{1}{(1-x)^2}=1+2x+3x^2+4x^3\dots(2)$

Now differentiate the given expression for $y_n(x)$ using the product rule. You will get a sum of n terms whose rth term is $rx^{r-1}y_n(x)$ and so

$\frac d{dx}\left(y_n(x)\right)=y_n(x)(1+2x+3x^2+\dots+nx^{n-1})$

Now use the laws of exponents to write $y_n(x)$ as a single exponential, and factor the exponent to show that this

$=e^{x(1+x+x^2+x^3+\dots+x^{n-1})}((1+2x+3x^2+\dots+nx^{n-1})$

Now let $n\rightarrow \infty$ and use (1) and (2) to write this as

$\frac {e^\frac{x}{1-x}}{(1-x)^2}$

Finally put $x=\frac1{2}$ and this simplifies easily to $\boxed{4e}$

First of all, observe that

$y_{n}(x) = e^{x+x^{2}+\cdots+x^{n}} = e^{\sum _{k=0}^{n}x^{k}}$

Then

$\frac{\mathrm{d} y_{n}(x)}{\mathrm{d} x} = \left [\sum _{j=1}^{n}jx^{j-1} \right ]e^{\sum _{k=0}^{n}x^{k}} = \left [ \frac{\mathrm{d} }{\mathrm{d} x}\sum _{j=0}^{n}x^{j} \right ] e^{\sum _{k=0}^{n}x^{k}}$

Therefore, taking the limit

$\lim _{n\rightarrow \infty} \frac{\mathrm{d} }{\mathrm{d} x}y_{n}(x) = \frac{1}{(1-x)^{2}} \; e^{\frac{1}{1-x}-1}$ $\therefore \left [\lim _{n\rightarrow \infty} \frac{\mathrm{d} }{\mathrm{d} x}y_{n}(x) \right ]_{x=\frac{1}{2}} = 4e$

$y_{n}\left(x\right)=\exp\left(x+x^{2}+x^{3}+\cdots+x^{n}\right)$ . Note that $\left(x+x^2+x^3+\cdots+x^n\right) \left(x-1\right) = \left(x^{n+1}-x\right)$ and therefore $y_{n}\left(x\right) = \exp\left(\frac{x^{n+1}-x}{x-1}\right).$ Applying the quotient rule we derive that $\frac{dy_{n}\left(x\right)}{dx}=\exp\left(\frac{x^{n+1}-x}{x-1}\right)\frac{nx^{n+1}-\left(n+1\right)x^{n}+1}{\left(x-1\right)^{2}}$ and therefore, $\left.\frac{dy_{n}\left(x\right)}{dx}\right|_{x=\frac{1}{2}}=\exp\left(1-2^{1-n}\right)\left(n2^{1-n}-\left(n+1\right)2^{2-n}+4\right)$ and as $n\rightarrow\infty$ we get $\lim_{n\rightarrow\infty}\left.\frac{dy_{n}\left(x\right)}{dx}\right|_{x=\frac{1}{2}}=4\exp\left(1\right)=4e$