<s>[INST] 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.
 [/INST] 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 $\displaystyle\lim_\limits{n\rightarrow\infty}\sum_\limits{k=1}^{n}\frac{k}{2^{k}}$.
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