25. Determine
$$
\lim _{n \rightarrow \infty} \sum_{i=0}^{n} \frac{1}{\binom{n}{i}} .
$$
(Note: Here $\binom{n}{i}$ denotes $\frac{n!}{i!(n-i)!}$ for $i=0,1,2,3, \cdots, n$.
25. Answer. 2
Solution. Let
$$
a_{n}=\sum_{i=0}^{n}\binom{n}{i}^{-1}
$$

Assume that $n \geq 3$. It is clear that
$$
a_{n}=2+\sum_{i=1}^{n-1}\binom{n}{i}^{-1}>2
$$

Also note that
$$
a_{n}=2+2 / n+\sum_{i=2}^{n-2}\binom{n}{i}^{-1}
$$

Since $\binom{n}{i} \geq\binom{ n}{2}$ for all $i$ with $2 \leq i \leq n-2$,
$$
a_{n} \leq 2+2 / n+(n-3)\binom{n}{2}^{-1} \leq 2+2 / n+2 / n=2+4 / n .
$$

So we have show that for all $n \geq 3$,
$$
2<a_{n} \leq 2+4 / n .
$$

Thus
$$
\lim _{n \rightarrow \infty} a_{n}=2
$$