Upper bound of $\sum_{j=1}^p \frac{p+1}{p-j+1} \frac1{2^j}$ I am looking for an upper bound of the following sum
$$
S_p:= \sum_{j=1}^p \frac{p+1}{p-j+1} \frac1{2^j}.
$$
The upper bound should be independent of $p$, of course. Numerical experiments indicate that 
$$
S_p \le \frac53
$$
with the maximum attained for $p=3,4$. However, I am not able to prove it.
I could only prove $S_p \le 3$. Here is, what I did:
First, we see
$$
\frac{p+1}{p-j+1} = 1 + \frac j{p-j+1}\le 1+j,
$$
hence
$$
S_p \le \sum_{j=1}^p  (1+j) \frac1{2^j} .
$$
Using the power series
$$
(1-x)^{-2} = \sum_{j=0}^\infty (1+j) x^j , \ |x|<1,
$$
I can estimate
$$
S_p \le (1-\frac12)^{-2} -1 = 3.
$$
How can the bound be improved?
 A: First, note that that the change of summation index $j\leftarrow p+1-j$ shows that
$$
S_p=\frac{p+1}{2^{p+1}}\sum_{j=1}^p\frac{2^j}{j}
$$
Thus
$$
\frac{2^{p+1}}{p+1}S_p=\sum_{j=1}^p\frac{2^j}{j}=\frac{2^p}{p}+\frac{2^{p}}{p}S_{p-1}
$$
Similarly$$\eqalign{
\frac{2^{p+2}}{p+2}S_{p+1} &=\frac{2^{p+1}}{p+1}+\frac{2^{p+1}}{p+1}S_{p}\cr
&=\frac{2p}{p+1}\left(\frac{2^{p+1}}{p+1}S_p-\frac{2^{p}}{p}S_{p-1}\right)+\frac{2^{p+1}}{p+1}S_{p}
}$$
This can be rearranged as follows
$$\eqalign{
\frac{S_{p+1}}{S_p} &= \frac{(p+2)(3p+1)}{2(p+1)^2}-\frac{p+2}{2(p+1)}\cdot\frac{S_{p-1}}{S_p}
}$$
Now, If for some $p\ge2$ we have $S_p<S_{p-1}$ then 
$$
\frac{S_{p+1}}{S_p}<\frac{(p+2)(3p+1)}{2(p+1)^2}-\frac{p+2}{2(p+1)}=\frac{p(p+2)}{(p+1)^2}<1.
$$ 
That is $S_{p+1}<S_p$, and we have proved that
$$
S_p<S_{p-1}\quad \Longrightarrow\quad S_{p+1}<S_p
$$
Now it is easy to check that $S_5=\frac{8}{5}<\frac{5}{3}=S_4$, hence the sequence
$(S_p)_p$ is decreasing starting from $p=4$. Since
$$S_1=1~<S_2=\frac{3}{2}<S_3=\frac{5}{3}=S_4=\frac{5}{3}$$
we conclude that
$$\max\{S_p:p\geq 1\}=\frac{5}{3}$$
and it is exactly attained at $p=3,4$.$\qquad\square$
Remark. as Did suggested, we have also the simple lower bound
$$
S_p\geq\frac{p+1}{2^{p+1}}\cdot\frac{2^{p+1}-2}{p}
$$
So, $\ell~{\buildrel{\rm def}\over =}~\lim\limits_{p\to\infty}S_p\geq 1$. This implies that
$$\inf\{S_p:p\geq 1\}=1=S_1.$$
Finally, with a little more work, it can be proved that $\ell=1$.
A: Hint: Rewrite the rational function $\frac{p+1}{p+1-j}$ as an infinite geometric series, i.e.,
$$\frac{p+1}{p+1-j}=\frac{1}{1-\left(\frac{j}{p+1}\right)}=\sum_{k=0}^{\infty}\left(\frac{j}{p+1}\right)^k.$$
This expansion is valid since the variable $j$ only ranges from $0\le j \le p<p+1$, hence $|\frac{j}{p+1}|<1.$
Then, by reversing the order of summation we get:
$$S_p=\sum_{j=1}^{p}\sum_{k=0}^{\infty}\left(\frac{j}{p+1}\right)^k\frac{1}{2^j}\\
=\sum_{k=0}^{\infty}\sum_{j=1}^{p}\left(\frac{j}{p+1}\right)^k\frac{1}{2^j}\\
=\sum_{k=0}^{\infty}\frac{1}{(p+1)^k}\sum_{j=1}^{p}\frac{j^k}{2^j}.$$
