Evaluate a limit in combinatorics

Question

Evaluate $$\lim_{n\rightarrow \infty} \sum_{r=0}^n \frac{1}{n\choose r}.$$

I tried it using the Reimann sum(turning it into a definite integral) but it got me confused with the factorials in it. I tried to replace factorials with the Gamma function but it got really messy!

I have a solution but don't know if its correct: N choose R represents number of ways to select R objects among N objects. So term in the summation is probability of selecting one particular objects of these objects. Now we are adding these probability because you might select the object alone or 2 at a time or 3 at a time and so on. So Integrand is probability of selecting an object which approaches zero as N approaches infinity.

Is this correct? and thanks in advance.

When $r=0$ or $r=n$ the associated summand is $1$, so the sum is always $\geq2$. — saulspatz, Sep 01 '21 at 19:12
The limit can't be zero, because all the terms are positive, and the terms are $1$ when $r=0$ and $r=n.$ — Thomas Andrews, Sep 01 '21 at 19:12

score 1 · Answer 1 · answered Sep 01 '21 at 19:33

Note that for $2 \leq r \leq n-2$, we have $$\frac{1}{\binom{n}{k}} \leq \frac{1}{\binom{n}{2}} = \frac{2}{n(n-1)}$$ So, $$ 2 \leq \sum_{k=0}^{n} \frac{1}{\binom{n}{k}} \leq 2 + \frac{2}{n} + \frac{2(n-3)}{n(n-1)}$$ Clearly, the term in the RHS goes to $2$ in the limit $n\to \infty$.

So, $$ \lim_{n\to\infty} \sum_{r=0}^{n} \frac{1}{\binom{n}{r}} = 2$$

Evaluate a limit in combinatorics

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