If $S_n=\displaystyle{\sum\limits_{r=0}^{n}\frac {1}{\binom{n}{r}}} $ and $T_n=\sum\limits_{r=0}^{n}\frac {r}{\binom{n}{r}}$ then what is $\displaystyle{\frac {T_n}{S_n}} $ in terms of $n$?
I tried the usual tricks I know from Binomial Theorem and Series but to no avail.
Can anyone help? Thanks
On
This is an extended comment showing (a) that closed expressions for $S$ and $T$ separately do exist, and (b) extends the relation to arbitrary functions of the binomial coefficient.
(a) closed expressions
These are not mentioned explicitly in OEIS. We designate them as $s_{c}$ and $t_{c}$, respectively.
Mathematica gives
$$\sum _{r=0}^n r! (n-r)! = 2^{-n-1} \Gamma (n+2) (-i\pi - B_2(n+2,0)) \tag{1}$$
where $B_z(a,b)$ is the incomplete Beta function (see e.g. http://mathworld.wolfram.com/IncompleteBetaFunction.html).
Hence follows
$$s_{c}(n) = 2^{-n-1} (n+1) (-B_2(n+2,0)-i \pi )$$
and, with the result $n/2$ for the quotient,
$$t_{c}(n) =\frac{n}{2} 2^{-n-1} (n+1) (-B_2(n+2,0)-i \pi )$$
The appearance of the imaginary term $-i \pi$ can be understood from the defining integral representation
$$B_z(a,b)= \int_0^z u^{a-1} (1-u)^{b-1} \, du\tag{2}$$
I find that there is simpler integral formula for $S$ where the singularity at $u= 1$ has to be treated with the principal value:
$$s_{cp}(n) = 2^{-n-1} (n+1) \;P\int_0^2 \frac{u^{n+1}}{u-1} \, du\tag{3}$$
(b) genralization to arbitrary functions
Due to the basic relation of the binomial coefficients
$$\binom{n}{r}=\binom{n}{n-r}$$
we have for any function f(x)
$$\sum _{r=0}^n r f(\binom{n}{r})=\sum _{r=0}^n (n-r) f(\binom{n}{n-r})=\sum _{r=0}^n (n-r) f(\binom{n}{r})=n \sum _{r=0}^n f(\binom{n}{r})-\sum _{r=0}^n f(\binom{n}{r}) $$
and it follows that
$$n/2 = \sum _{r=0}^n r f(\binom{n}{r})/\sum _{r=0}^n f(\binom{n}{r})\tag{4}$$
Use the symmetry of the binomial coefficients: $\binom{n}{r}=\binom{n}{n-r}$. Then $$\sum_{r=0}^n \frac{r}{\binom{n}{r}}=\sum_{r=0}^n \frac{n-r}{\binom{n}{n-r}}=\sum_{r=0}^n \frac{n-r}{\binom{n}{r}}=n\sum_{r=0}^n \frac{1}{\binom{n}{r}}-\sum_{r=0}^n \frac{r}{\binom{n}{r}}.$$ Can you take it from here?