Limit involving binomial coefficient: $\lim_{n\to\infty} \left(\frac n{n-1}\right)^2 \left(\frac12\right)^n \sum_{i=1}^n \binom ni \cdot \frac{i-1}i$

$$\sum_{i=1}^{n}\binom{n}{i}\frac{i-1}{i} = 2^n-1-\sum_{i=1}^{n}\binom{n}{i}\frac{1}{i}$$ and: $$\begin{eqnarray*}\sum_{i=1}^{n}\binom{n}{i}\frac{1}{i}&=&\sum_{i=1}^{n}\binom{n}{i}\int_{0}^{1}x^{i-1}\,dx = \int_{0}^{1}\frac{(1+x)^n-1}{x}\,dx\\&=&\int_{0}^{1}\frac{(2-x)^n-1}{1-x}\,dx=\sum_{j=0}^{n-1}\int_{0}^{1}(2-x)^j\,dx\\&=&\sum_{j=0}^{n-1}\frac{2^{j+1}-1}{j+1}=\sum_{i=1}^{n}\frac{2^i-1}{i}=\sum_{i=1}^{n}\frac{2^i}{i}-H_n \end{eqnarray*}$$ so your limit is: $$ \lim_{n\to +\infty}\left(\frac{n}{n-1}\right)^2\cdot\left(1-\frac{H_n+1}{2^n}-\sum_{i=0}^{n-1}\frac{1}{(n-i) 2^{i}}\right)=\color{red}{1}$$ since: $$ \sum_{i=0}^{n-1}\frac{1}{(n-i)2^i}\leq\sum_{i=0}^{n-1}\frac{1}{(n-i)(1+i)}=\frac{1}{n+1}\sum_{i=0}^{n-1}\left(\frac{1}{i+1}+\frac{1}{n-i}\right)=\frac{2H_n}{n+1}.$$

Define $$ \begin{align} S_n &=\sum_{k=1}^n\binom{n}{k}\frac1k\\ &=\sum_{k=1}^n\left[\binom{n-1}{k-1}+\binom{n-1}{k}\right]\frac1k\\ &=\frac1n\sum_{k=1}^n\binom{n}{k}+S_{n-1}\\ &=\frac{2^n-1}n+S_{n-1} \end{align} $$ Therefore, $$ \sum_{k=1}^n\binom{n}{k}\frac1k=\sum_{k=1}^n\frac{2^k-1}k $$ Note that $$ \begin{align} \lim_{n\to\infty}\frac n{2^n}\sum_{k=1}^n\frac1k &\le\lim_{n\to\infty}\frac{n^2}{2^n}\\ &=0 \end{align} $$ and by Dominated Convergence $$ \begin{align} \lim_{n\to\infty}\frac{n}{2^n}\sum_{k=1}^n\frac{2^k}k &=\lim_{n\to\infty}\frac{n}{2^n}\sum_{k=0}^{n-1}\frac{2^{n-k}}{n-k}\\ &=\lim_{n\to\infty}\sum_{k=0}^{n-1}2^{-k}+\lim_{n\to\infty}\sum_{k=0}^{n-1}\frac{k\,2^{-k}}{n-k}\\[6pt] &=2+0 \end{align} $$ Therefore, $$ \begin{align} \lim_{n\to\infty}\frac1{2^n}\sum_{k=1}^n\binom{n}{k}\left(1-\frac1k\right) &=\lim_{n\to\infty}\left(1-\frac1{2^n}-\frac2n\right)\\ &=1 \end{align} $$