Is this limit solvable using Stolz’s theorem ? $\lim_\limits{n\to+\infty}\left(\frac12+\frac3{2^2}+\dots+\frac{2n−1}{2^n}\right)$ I would like to ask if this example is solvable using Stolz’s theorems, because I have a $2^n$ expression at the bottom and a quadratic expression at the top, so it doesn't work for me, but maybe I'm calculating wrong.
 A: Another way to calculate the limit $\lim_\limits{n\to+\infty}\!\left(\frac12\!+\!\frac3{2^2}\!+\!\ldots\!+\!\frac{2n−1}{2^n}\right).$
Let $\;a_n=\dfrac12+\dfrac3{2^2}+\dots+\dfrac{2n−1}{2^n}\;\;$ for any $\;n\in\mathbb N\,.$
It results that $\;a_{n+1}-a_n=\dfrac{2n+1}{2^{n+1}}\;\;$ for any $\;n\in\mathbb N\,.$
Now, I am going to look for another sequence $\big\{b_n\big\}_{\!n\in\mathbb N}$ such that $\;b_{n+1}-b_n=\dfrac{2n+1}{2^{n+1}}\;\;$ for any $\;n\in\mathbb N\,.\quad\color{blue}{(1)}$
By letting $\;c_n=2^nb_n\;,\;$ the equality $\;(1)\;$ turns to :
$c_{n+1}-2c_n=2n+1\;\;$ for any $\;n\in\mathbb N\,.\quad\color{blue}{(2)}$
Moreover, by letting $\;d_n=2n+c_n\;,\;$ the equality $(2)$ turns to :
$d_{n+1}-2d_n=3\;\;$ for any $\;n\in\mathbb N\,.\quad\color{blue}{(3)}$
The sequence , $\;d_n=-3\;$ for all $\;n\in\mathbb N\;,\;$ satisfies $\;(3)\;,\;$ consequently ,
the sequence , $\;c_n=-2n-3\;$ for all $\;n\in\mathbb N\;,\;$ satisfies $\;(2)\;$ and the sequence , $\;b_n=\dfrac{-2n-3}{2^n}\;$ for all $\;n\in\mathbb N\;,\;$ satisfies the equality $\;(1)\;.$
Since $\;a_{n+1}-a_n=b_{n+1}-b_n\;\;$ for all $\;n\in\mathbb N\;,\;$ it follows that
$a_{n+1}-b_{n+1}=a_n-b_n\;\;$ for all $\;n\in\mathbb N\;,\;$ hence ,
the sequence $\big\{a_n-b_n\big\}_{\!n\in\mathbb N}\,$ is constant , consequently ,
$a_n-b_n=a_1-b_1=\dfrac12+\dfrac52=3\;\;$ for all $\;n\in\mathbb N\;,$
$a_n=3+b_n\;\;$ for all $\;n\in\mathbb N\;,$
$a_n=3-\dfrac{2n+3}{2^n}\;\;$ for all $\;n\in\mathbb N\;.$
Therefore ,
$\exists\lim_\limits{n\to+\infty}a_n=3\;.$
A: Denote $S_{n}=\frac{1}{2}+\frac{3}{2^{2}}+\ldots+\frac{2n-1}{2^{n}}$.
Firstly, observe that
\begin{eqnarray*}
S_{n} & = & \left\{ \frac{2}{2}+\frac{4}{2^{2}}+\frac{2n}{2^{n}}\right\} -\left\{ \frac{1}{2}+\frac{1}{2^{2}}+\ldots+\frac{1}{2^{n}}\right\} \\
 & = & 2\sum_{k=1}^{n}\frac{k}{2^{k}}-\frac{\frac{1}{2}\left[1-(\frac{1}{2})^{n}\right]}{1-\frac{1}{2}}\\
 & = & 2\sum_{k=1}^{n}\frac{k}{2^{k}}-\left\{ 1-\left(\frac{1}{2}\right)^{n}\right\} .
\end{eqnarray*}
Now, we focus on the computation of $\sum_{k=1}^{n}\frac{k}{2^{k}}$.
Let $f(x)=\sum_{k=1}^{n}x^{k}$, then $f'(x)=\sum_{k=1}^{n}kx^{k-1}$.
Therefore, $\sum_{k=1}^{n}kx^{k}=xf'(x)$. On the other hand, by sum
of G.P. formula (which is valid for $x\neq 1$), $f(x)=\frac{x(1-x^{n})}{1-x}$, so
\begin{eqnarray*}
f'(x) & = & \frac{\left(1-(n+1)x^{n}\right)(1-x)+x(1-x^{n})}{(1-x)^{2}}.
\end{eqnarray*}
Hence, we obtain the formula:
\begin{eqnarray*}
\sum_{k=1}^{n}kx^{k} & = & x\cdot\frac{\left(1-(n+1)x^{n}\right)(1-x)+x(1-x^{n})}{(1-x)^{2}}.
\end{eqnarray*}
Put $x=\frac{1}{2}$, then we obtain
$$
\sum_{k=1}^{n}\frac{k}{2^{k}}=\frac{1}{2}\cdot\frac{\left[1-(n+1)\left(\frac{1}{2}\right)^{n}\right]\cdot\frac{1}{2}+\frac{1}{2}\left[1-\left(\frac{1}{2}\right)^{n}\right]}{\frac{1}{4}}.
$$
Now, you should be able to compute the limit $\displaystyle\lim_\limits{n\rightarrow\infty}\sum_\limits{k=1}^{n}\frac{k}{2^{k}}$.
