Determine the value of the sum to infinity of $(u_{r+1} + \frac{1}{2^r})$ 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?

Write $$U_{r+1}=\frac{2}{(r+2)(r+4)}= 2 \left[\left(\frac{1}{r+2}-\frac{1}{r+3}\right)+\left(\frac{1}{r+3}-\frac{1}{r+4}\right) \right]= [(F_r-F_{r+1})+(F_{r+1}-F_{r+2})], F_r=\frac{1}{r+2}.$$
Now telescoping summation can be done easily.
