Calculate sums of inverses of binomial coefficients How to calculate the sum of sequence $$\frac{1}{\binom{n}{1}}+\frac{1}{\binom{n}{2}}+\frac{1}{\binom{n}{3}}+\cdots+\frac{1}{\binom{n}{n}}=？$$ How about its limit?
 A: Here is a method that I just came up with in chat
$$
\begin{align}
\frac1{\binom{n}{k\vphantom{+1}}}&=\frac{n-k}{n}\frac1{\binom{n-1}{k}}\tag{1}\\
\frac1{\binom{n}{k+1}}&=\frac{k+1}{n}\frac1{\binom{n-1}{k}}\tag{2}\\
\sum_{k=0}^{n-1}\left(\frac1{\binom{n}{k\vphantom{+1}}}+\frac1{\binom{n}{k+1}}\right)
&=\frac{n+1}{n}\sum_{k=0}^{n-1}\frac1{\binom{n-1}{k}}\tag{3}\\
2\sum_{k=0}^n\frac1{\binom{n}{k\vphantom{+1}}}
&=2+\frac{n+1}{n}\sum_{k=0}^{n-1}\frac1{\binom{n-1}{k}}\tag{4}\\
\frac{2^{n+1}}{n+1}\sum_{k=0}^n\frac1{\binom{n}{k\vphantom{+1}}}
&=\frac{2^{n+1}}{n+1}+\frac{2^n}{n}\sum_{k=0}^{n-1}\frac1{\binom{n-1}{k}}\tag{5}\\
\frac{2^{n+1}}{n+1}\sum_{k=0}^n\frac1{\binom{n}{k\vphantom{+1}}}
&=\sum_{k=1}^{n+1}\frac{2^k}{k}\tag{6}\\
\sum_{k=0}^n\frac1{\binom{n}{k\vphantom{+1}}}
&=\frac{n+1}{2^{n+1}}\sum_{k=1}^{n+1}\frac{2^k}{k}\tag{7}\\
\end{align}
$$
Explanation
$(1)$: Binomial identity: $\frac{n!}{k!(n-k)!}=\frac{n}{n-k}\frac{(n-1)!}{k!(n-k-1)!}$
$(2)$: Binomial identity: $\frac{n!}{(k+1)!(n-k-1)!}=\frac{n}{k+1}\frac{(n-1)!}{k!(n-k-1)!}$
$(3)$: Add $(1)$ and $(2)$ and sum $\vphantom{\frac{()}{()}}$
$(4)$: Add $\frac1{\binom{n}{n\vphantom{+1}}}+\frac1{\binom{n}{0}}=2$ to both sides
$(5)$: multiply both sides by $\frac{2^n}{n+1}$
$(6)$: $a_n=\frac{2^{n+1}}{n+1}+a_{n-1}$ where $a_n=\frac{2^{n+1}}{n+1}\sum\limits_{k=0}^n\frac1{\binom{n}{k\vphantom{+1}}}$
$(7)$: multiply both sides by $\frac{n+1}{2^{n+1}}$

Limit
For $2\le k\le n-2$, we have that $\binom{n}{k}\ge\binom{n}{2}$. Thus, for $n\ge4$,
$$
\begin{align}
\sum_{k=0}^n\frac1{\binom{n}{k}}
&=\overset{\substack{k=0\\k=n\\\downarrow\\[3pt]\,}}{2\vphantom{\frac2n}}+\overset{\substack{k=1\\k=n-1\\\downarrow\\[3pt]\,}}{\frac2n}+\sum_{k=2}^{n-2}\frac1{\binom{n}{k}}\tag8\\
&\le2+\frac2n+\frac{n-3}{\binom{n}{2}}\tag9\\
&\le2+\frac4n\tag{10}
\end{align}
$$
Therefore, for $n\ge4$,
$$
2+\frac2n\le\sum_{k=0}^n\frac1{\binom{n}{k}}\le2+\frac4n\tag{11}
$$
and the Squeeze Theorem says
$$
\lim_{n\to\infty}\sum_{k=0}^n\frac1{\binom{n}{k}}=2\tag{12}
$$
A: When interested in the limit only, just observe that for $2 \leq k \leq n-2$, we have
$$\frac{1}{\binom{n}{k}} \leq \frac{1}{\binom{n}{2}} = \frac{2}{n(n-1)}.$$
Thus
$$ 2 \leq \sum_{k=0}^{n} \frac{1}{\binom{n}{k}} \leq 2 + \frac{2}{n} + \frac{2(n-3)}{n(n-1)} \xrightarrow[n\to\infty]{} 2$$
and therefore
$$ \lim_{n\to\infty} \sum_{k=0}^{n} \frac{1}{\binom{n}{k}} = 2.$$
A: Assuming $C_n^k$ stands for $\binom{n}{k} = \frac{n!}{(n-k)! \cdot k!}$, and using, for $k>0$: 
$$
   \frac{1}{\binom{n}{k}} = k \operatorname{Beta}(k,n-k+1) = k \int_0^1 (1-x)^{k-1} x^{n-k} \mathrm{d} x
$$
Hence
$$
   S = \sum_{k=1}^n \frac{1}{\binom{n}{k}} =  \int_0^1 \sum_{k=0}^n k (1-x)^{k-1} x^{n-k} \mathrm{d} x = \int_0^1 \frac{x^{n+1} -(1-x)^n ((2n+1)x-n)}{(1-2x)^2} \mathrm{d} x
$$
Now changing integration variable $x = \frac{1}{2} + u$:
$$\begin{eqnarray}
 S &=& \int_{-1/2}^{1/2} \frac{\mathrm{d} u}{4 u^2} \left( \left(\frac{1}{2}+u\right)^{n+1} - \frac{1}{2} \left(\frac{1}{2}-u\right)^n \left( 1 + 2 (2n+1) u\right)\right) \\
   &=& \int_{-1/2}^{1/2} \frac{\mathrm{d} u}{4 u^2} \sum_{k=2}^{n+1} 2^{k-n-1} \left((-1)^k \left((2 n+1) \binom{n}{k-1}-\binom{n}{k}\right)+\binom{n+1}{k}\right) u^{k} \\
   &=&   \sum_{k=1}^{\lfloor\frac{n+1}{2}\rfloor} 2^{2k-n-1} \left(\left((2 n+1) \binom{n}{2k-1}-\binom{n}{2k}\right)+\binom{n+1}{2k}\right) \underbrace{\int_{-1/2}^{1/2} \frac{\mathrm{d} u}{4 u^2} u^{2 k}}_{\frac{1}{4^k} \frac{1}{2k-1}} \\
 &=& \sum_{k=1}^{\lfloor\frac{n+1}{2}\rfloor} \frac{1}{2^{n}} \frac{1}{2k-1}\frac{(n+1)!}{ (2k-1)! (n-2k+1)!} \stackrel{\ast}{=}  \sum_{k=1}^n \frac{n+1}{n+1-k} \frac{1}{2^k}
\end{eqnarray}
$$
Thus
$$
   \sum_{k=0}^n \frac{1}{\binom{n}{k}} = \sum_{k=0}^n \frac{n+1}{n+1-k} \frac{1}{2^k}
$$
For large $n$ the sum approaches the value of $2$ from above:

I am hoping this sum has a nice probabilistic underpinnings to it.

Added: Derivation of equality $\stackrel{\ast}{=}$ above
$$ \begin{eqnarray}
  S &=&  \sum_{k=1}^{\lfloor\frac{n+1}{2}\rfloor} \frac{1}{2^{n}} \frac{1}{2k-1}\frac{(n+1)!}{ (2k-1)! (n-2k+1)!} = \frac{n+1}{2^n} \underbrace{\sum_{k=0}^{\lfloor\frac{n-1}{2}\rfloor} \frac{1}{2k+1} \binom{n}{2k+1}}_{A_n} 
\end{eqnarray}
$$
We will now establish that $A_{n+1} - A_{n} = \frac{2^{n}}{n+1}$, which would imply $$A_n = \sum_{k=1}^n \frac{2^{k-1}}{k} \stackrel{k \to n+1-k}{=} \sum_{k=1}^n \frac{2^{n-k}}{n+1-k}$$
We thus proceed, firstly for even, $n=2m$:
$$ \begin{eqnarray}
   A_{2m+1} -A_{2m} &=& \sum_{k=0}^{m} \frac{1}{2k+1} \left( \underbrace{\binom{2m+1}{2k+1} - \binom{2m}{2k+1}}_{\binom{2m}{2k}}\right)  \\
  &=& \sum_{k=0}^{m} \frac{1}{2k+1} \binom{2m}{2k} = \frac{1}{2}\int_0^1 \left( (1+x)^{2m} + (1-x)^{2m} \right) \mathrm{d} x = \frac{2^{n}}{n+1}
\end{eqnarray}
$$
and then, similarly, for odd, $n=2m+1$:
$$\begin{eqnarray}
  A_{2m+2}-A_{2m+1} &=& \sum_{k=0}^m \frac{1}{2k+1} \binom{2m+1}{2k} \\
    &=& \frac{1}{2}\int_0^1 \left( (1+x)^{2m+1} + (1-x)^{2m+1} \right) \mathrm{d} x = \frac{2^n}{n+1}
\end{eqnarray}
$$
A: The determination of the limit is direct, keeping only the first and last terms and bounding the others. To get an exact formula, one can use a method similar to @Sasha's while (i) being somewhat simpler and (ii) avoiding a step I find unclear. Like @Sasha, one starts with a beta representation, namely,
$$
{n\choose k}^{-1}=(n+1)\int_0^1x^{n-k}(1-x)^k\mathrm dx.
$$
Summing up,
$$
S_n=\sum_{k=0}^n{n\choose k}^{-1}=(n+1)\int_0^1u_n(x)\mathrm dx,\quad u_n(x)=\sum_{k=0}^nx^{n-k}(1-x)^k.
$$
Note that $u_n(x)$ is a geometric series, hence
$$
u_n(x)=\frac{x^{n+1}-(1-x)^{n+1}}{2x-1}.
$$
Using the change of variables $x=\frac12(1+z)$ with $-1\leqslant z\leqslant 1$ yields
$$
u_n(x)=\frac{(1+z)^{n+1}-(1-z)^{n+1}}{2^{n+1}z}=\frac1{2^{n}}\sum_k{n+1\choose 2k+1}z^{2k}.
$$
Thus,
$$
S_n=\frac{n+1}{2^n}\sum_k{n+1\choose 2k+1}\frac1{2k+1}\left[z^{2k+1}\right]_{0}^1,
$$
and finally
$$
\sum_{k=1}^n{n\choose k}^{-1}=S_n-1=\frac1{2^n}\sum_{k=0}^{\lfloor\frac{n}2\rfloor}{n+1\choose 2k+1}\frac{n+1}{2k+1}-1.
$$
A: As an alternative, starting from this result by robjohn's answer [tag (7)]
$$\sum_{k=0}^n\frac1{\binom{n}{k}}=\frac{n+1}{2^{n+1}}\sum_{k=1}^{n+1}\frac{2^k}{k}=\frac{\sum_{k=1}^{n+1}\frac{2^k}{k}}{\frac{2^{n+1}}{n+1}}$$
by Stolz-Cesaro we obtain
$$\frac{\frac{2^{n+2}}{n+2}}{\frac{2^{n+2}}{n+2}-\frac{2^{n+1}}{n+1}}=\frac{2}{2-\frac{n+2}{n+1}}\to 2$$
