Solving a limit involving binomial coefficients

Asked Apr 13 '22 at 03:49

Active Apr 13 '22 at 03:49

Viewed 33 times

If $$a_n=\sum_{i=0}^{n}\frac{1}{n\choose i}$$ then find $$L=\lim_{n \to \infty}\frac{1}{a_n}$$

I tried to solve it by using sandwich theorem and obtained following relation

$$\frac{1}{\frac{1}{n\choose 0}+\frac{1}{n\choose n}}\geq \frac{1}{a_n}\geq0$$

The first inequality is obtained using the fact that $$\frac{1}{n\choose 0}+\frac{1}{n\choose n}\leq \sum_{i=0}^{n}\frac{1}{n\choose i}$$

This gives $$0.5\geq L \geq 0$$

The answer is $0.5$ only , however I am not able to prove that $$\frac{1}{a_n}\geq 0.5$$

How to continue?

asked Apr 13 '22 at 03:49

Lalit Tolani

Hint: $a_n\geq \frac{n}{{n \choose \lfloor\frac{n}{2}\rfloor}}$. – AnCar Apr 13 '22 at 04:04
You already showed $0.5 \ge 1/a_n$ for each $n \ge 2.$ If you were to show the reverse inequality, it would mean that $1/a_n$ (and thus also $a_n$) was constant for $n \ge 2.$ But it is not. – coffeemath Apr 13 '22 at 04:08
See this answer obtained using the formula searching tool https://approach0.xyz/ – Jean Marie Apr 13 '22 at 04:15
@AnCar I am not able to see how does that helps – Lalit Tolani Apr 13 '22 at 05:19
@coffeemath Hmm, Yes you are right , I need to look for some $f(n)$ such that $\frac{1}{a_n}\geq f(n)$ and $\lim_{n\to\infty}f(n)=0.5$ – Lalit Tolani Apr 13 '22 at 05:22
@JeanMarie Yes the answer helps me, thanks for helping :-) – Lalit Tolani Apr 13 '22 at 05:30

0 Answers0