Determine the value of $$\sum_{r=1}^\infty\Bigl(u_{r+1} + \frac{1}{2^r}\Bigr)$$.
In earlier parts of the question, it is given $\displaystyle u_r= \frac{2}{(r+1)(r+3)}$, which when expressed in partial fractions is $\displaystyle\frac{1}{r+1}-\frac{1}{r+3}$.
Also previously, the question asks to show that $$\sum_{r=1}^n u_r = \frac{5}{6}-\frac {1}{n+2} -\frac{1}{n+3}$$ which I managed to.
I have also determined the values of $$\sum_{r=1}^\infty u_r$$ which is $$\sum_{r=1}^\infty u_r= \frac{5}{6}$$
But I don’t know how to solve for
$$\sum_{r=1}^\infty\Bigl(u_{r+1} + \frac{1}{2^r}\Bigr)$$
My attempts so far:
$\displaystyle u_{r+1} =\frac{2}{(r+1+1)(r+1+3)}$
$=\displaystyle\frac{2}{(r+2)(r+4)}$
Let $\displaystyle \frac{2}{(r+2)(r+4)}\equiv \frac{A}{(r+2)(r+4)}$
$2\equiv A(r+4)+B(r+2)$
Let r $=-4,B=-1$
Let $r=-2,A=1$
So, is $\displaystyle u_{r+1} \equiv \frac {1}{r+2}- \frac{1}{r+4}$?
Assuming that it is,$$\sum_{r=1}^\infty\Bigl(u_{r+1} + \frac{1}{2^r}\Bigr)$$
= $$\sum_{r=1}^\infty\Bigl(\frac{1}{r+2}+\frac{1}{r+4} + \frac{1}{2^r}\Bigr)$$
$=\frac{1}{3}-\require{cancel}\cancel{\frac{1}{5}}+\frac{1}{2}$
$\frac{1}{4}-\require{cancel}\cancel{\frac{1}{6}}+\frac{1}{4}$
$\require{cancel}\cancel{\frac{1}{5}}-\cancel{\frac{1}{7}}+\frac{1}{8}$
$\require{cancel}\cancel{\frac{1}{6}}-\cancel{\frac{1}{8}}+\frac{1}{16}$
.....
I don't know how to cancel the middle terms, how should this question be solved?
You have
$$\begin{equation}\begin{aligned} \sum_{r=1}^\infty\Bigl(u_{r+1} + \frac{1}{2^r}\Bigr) & = \sum_{r=1}^\infty u_{r+1} + \sum_{r=1}^\infty\frac{1}{2^r} \\ & = (\sum_{r=1}^\infty u_{r} - u_1) + 1 \\ & = \frac{5}{6} -\frac{1}{4} + 1 \\ & = \frac{10}{12} - \frac{3}{12} + \frac{12}{12} \\ & = \frac{19}{12} \end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Note I used $\sum_{r=1}^\infty\frac{1}{2^r}$ is the sum of an infinite geometric series, with $a = \frac{1}{2}$ being the first term and $r = \frac{1}{2}$ being the common ratio, so the sum is $\frac{a}{1-r} = 1$.