If $\ U_{r} = \frac{1+\ U_{r-1}}{2}$ and $\ U_{0}=0$, Find $\lim_{n\to\infty} \sum_{r=1}^n \ U_{r}$

Question

I am trying to understand fully how drug half-life works. So I derived this relationship:

$$\ U_{r} = \frac{1+\ U_{r-1}}{2}$$ Where $\ U_{0}=0$ and r is a set of natural numbers.

My issue to how to deduce a relationship for the sum to infinity:

$$\ S_{\infty}=\lim_{n\to\infty} \sum_{r=1}^n \ U_{r}$$

Consequently I need to get the relationship for $\ S_{\infty}$ if $\ U_{r} = \frac{A+\ U_{r-1}}{2}$ and $\ U_{0}=0$

Answer 1

$U_r=\frac{1}{2}+\frac{U_{r-1}}{2}=\frac{1}{2}+\frac{1}{4}+\frac{U_{r-2}}{4}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{U_{r-3}}{8}=...=\sum_{i=1}^{r}\frac{1}{2^i}+\frac{U_0}{2^r}=1-\frac{1}{2^r}$

and hence $\sum_{1}^{n}U_r=n-\sum_1^n\frac{1}{2^r}$

Answer 2

Suppose $u_0,u_1,u_2,...$ are defined by $$ \left\lbrace \begin{align*} u_0\!&=\,0\\[4pt] u_{n+1}\!&=\frac{a+u_n}{2}\\[4pt] \end{align*} \right. $$ for some constant $a > 0$.

Then we have \begin{align*} u_1&=\frac{a}{2}+\frac{1}{2}u_0\\[4pt] u_2&=\frac{a}{2}+\frac{1}{2}u_1\\[4pt] u_3&=\frac{a}{2}+\frac{1}{2}u_2\\[4pt] \!\!\!\!\!\vdots&\phantom{=\frac{a}{2}}\;\;\,\vdots\\[4pt] u_{n}&=\frac{a}{2}+\frac{1}{2}u_{n-1}\\[4pt] \end{align*} Summing the above equations, we get $$\;\;\;\;S_n=\frac{a}{2}{\,\cdot\,}\,n+\frac{1}{2}S_{n-1}$$ but it's easily shown that for all $k$ we have $S_k\ge 0$, so $$S_n\ge \frac{a}{2}{\,\cdot\,}\,n\qquad\;\;$$ hence ${\displaystyle{\lim_{n\to\infty}}}S_n=\infty$.