$a_k=(\frac{1}{2})^k-\frac{k}{n-k+1}a_{k-1}$ and $a_1=\frac{1}{2}+\frac{1}{n}(\frac{1}{2})^n-\frac{1}{n}$.($n\ge2$).I want to compute $\lim_{n\to\infty}\sum_{k=1}^{n-1}a_k$.
My first attempt is to find out a general form for $a_k$, and I tried to cast the recurrence relation into the form $f(k)a_k+f(k-1)a_{k-1}=g(k)$ and then the recurrence relation will become linear constant-coefficient one by taking $b_k=f(k)a_k$. However, I failed to convert it.
Also, since $\lim_{n\to\infty}\sum_{k=1}^{n-1}a_k\le\sum_{k=1}^{\infty}(\frac{1}{2})^k=1$, I tried to see if we can evaluate $\lim_{n\to\infty}\sum_{k=1}^{n-1}\frac{k}{n-k+1}a_{k-1}$.
Thanks for any hint.
Preliminary Result
Using the Beta Function Integral, we get $$ \begin{align} \sum_{k=0}^m\frac{(-1)^k}{\binom{n}{k}} &=(n+1)\sum_{k=0}^m\int_0^1(-1)^kt^k(1-t)^{n-k}\,\mathrm{d}t\\ &=(n+1)\sum_{k=0}^m\int_0^1(1-t)^n\left(\frac t{t-1}\right)^k\,\mathrm{d}t\\ &=(n+1)\int_0^1(1-t)^{n+1}\left(1-\left(\frac t{t-1}\right)^{m+1}\right)\mathrm{d}t\\ &=(n+1)\left(\frac1{n+2}+\frac{(-1)^m}{n+2}\frac{(n-m)!(m+1)!}{(n+1)!}\right)\\ &=\frac{n+1}{n+2}\left[1+\frac{(-1)^m}{\binom{n+1}{m+1}}\right]\tag{1} \end{align} $$ Now that we have this result, we could verify it more easily by induction.
Apply To The Question
Start with $$ a_k=\frac1{2^k}-\frac{k}{n-k+1}a_{k-1}\tag{2} $$ and $$ a_1=\frac12+\frac1n\frac1{2^n}-\frac1n\tag{3} $$ Multiply $(2)$ by $(-1)^k\binom{n}{k}$ and rearrange $$ \underbrace{(-1)^k\binom{n}{k}a_k}_{\large u_k}-\underbrace{(-1)^{k-1}\binom{n}{k-1}a_{k-1}}_{\large u_{k-1}}=(-1)^k\binom{n}{k}\frac1{2^k}\tag{4} $$ Thus, we can solve for $u_k=(-1)^k\binom{n}{k}a_k$ $$ \begin{align} \overbrace{(-1)^k\binom{n}{k}a_k}^{\large u_k} &=\overbrace{\ -na_1\ \ \vphantom{\binom{n}{j}}}^{\large u_1}+\sum_{j=2}^k\overbrace{(-1)^j\binom{n}{j}\frac1{2^j}}^{\large u_j-u_{j-1}}\\ &=-\frac1{2^n}+\sum_{j=0}^k(-1)^j\binom{n}{j}\frac1{2^j}\tag{5} \end{align} $$ Therefore, $$ \begin{align} \sum_{k=0}^na_k &=-\frac1{2^n}\sum_{k=0}^n\frac{(-1)^k}{\binom{n}{k}}+\sum_{j=0}^n\sum_{k=j}^n\frac{(-1)^{j+k}}{2^j}\frac{\binom{n}{j}}{\binom{n}{k}}\tag{6a}\\ &=-\frac1{2^n}\frac{n+1}{n+2}\left(1+(-1)^n\right)+\frac{n+1}{n+2}\sum_{j=0}^n\left(-\frac12\right)^j\binom{n}{j}\left[(-1)^n+\frac{(-1)^j}{\binom{n+1}{j}}\right]\tag{6b}\\ &=-\frac1{2^n}\frac{n+1}{n+2}\left(1+(-1)^n\right)+\left(-\frac12\right)^n\frac{n+1}{n+2}\\ &+\frac{n+1}{n+2}\sum_{j=0}^n\frac1{2^j}\left(1-\frac j{n+1}\right)\tag{6c}\\ &=-\frac1{2^n}\frac{n+1}{n+2}\left(1+(-1)^n\right)+\left(-\frac12\right)^n\frac{n+1}{n+2}\\ &+\frac{n+1}{n+2}\left(2-\frac1{2^n}\right)-\frac1{n+2}\left(2-\frac{n+2}{2^n}\right)\tag{6d}\\ &=\frac{n}{n+2}\left(2-\frac1{2^n}\right)\tag{6e} \end{align} $$ Explanation:
$\text{(6a)}$: multiply $(5)$ by $\left.(-1)^k\middle/\binom{n}{k}\right.$ and sum in $k$, then change the order of summation
$\text{(6b)}$: apply $(1)$
$\text{(6c)}$: Binomial Theorem and $\left.\binom{n}{j}\middle/\binom{n+1}{j}\right.=1-\frac j{n+1}$
$\text{(6d)}$: $\sum\limits_{j=0}^n\frac1{2^j}=2-\frac1{2^n}$ and $\sum\limits_{j=0}^n\frac j{2^j}=2-\frac{n+2}{2^n}$
$\text{(6e)}$: combine terms
Subtracting $a_0=1-\frac1{2^n}$ and $a_n=0$ from $\text{(6e)}$, we get $$ \bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^{n-1}a_k =\frac{n-2}{n+2}+\frac2{n+2}\frac1{2^n}}\tag{7} $$ and finally, $$ \bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}\sum_{k=1}^{n-1}a_k=1}\tag{8} $$