<s>[INST] 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?
 [/INST] 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
$$\frac{\frac{2^{n+2}}{n+2}}{\frac{2^{n+2}}{n+2}-\frac{2^{n+1}}{n+1}}=\frac{2}{2-\frac{n+2}{n+1}}\to 2$$
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