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I'm in a calculus class I need to prove that $$\lim_{n\to\infty}\sum_{k=0}^{n}\frac{1}{{n\choose k}} \to 2$$ Is my below solution valid?

We are given the problem

$$\lim_{n\to\infty}\sum_{k=0}^{n}\frac{1}{{n\choose k}}$$ Looking at this, we can see that the sum is always at least $2$, for: $$\frac{1}{{n \choose 0}} + \frac{1}{{n \choose n}} = 1+1$$ $$= 2$$ looking at the above, we see that each term must be $\le 1$ for ${n \choose k}$ must always be less than or equal to one. Therefore:

$$\sum_{k=0}^{n}\frac{1}{{n\choose k}} \le n \cdot 1 = n$$

From the above, we can conclude that the graph of the function looks like an upside-down normal distribution curve, this is because it starts at 1, decreases, and then increases back to 1. From this, we can see that the function is symmetrical, and thus all the numbers in the middle(not including the edge cases) must be less than:

$$\frac{1}{{n \choose 1}} = \frac{1}{{n \choose n-1}} = \frac{1}{n}$$ Therefore, since $$\lim_{n\to\infty}\frac{1}{n} \to 0$$ Our final answer is $$1+0+0+....+0+0+1 = 2$$ And therefore, $$\lim_{n\to\infty}\sum_{k=0}^{n}\frac{1}{{n\choose k}} \to 2$$

Omori
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    If this reasoning was correct, all Riemann integrals would be zero. The individualterms go to zero, but the number of terms goes to infinity, so it is possible for the sequence to converge to a non-zero value. – Thomas Andrews Nov 09 '23 at 19:54

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"[W]e can conclude that the graph of the function looks like an upside-down normal distribution curve, this is because it starts at 1, decreases, and then increases back to 1" is not properly justified (though it is a true statement that we can bound $\frac{1}{\binom{n}{k}}$ by $\frac{1}{\binom{n}{1}}$ for $1 \leq k \leq n-1$).

It's true that the limit of a sum is the sum of the limits, but if a sum contains a variable number of elements, this no longer holds. Note that by your reasoning, the sum of $\frac{1}{n}$ with itself $n$ times would limit to $0$, when for any $n$, it equals $1$. (This mistake feels similar in flavor to me as this argument Where is the flaw in this "proof" that 1=2? (Derivative of repeated addition)) To prove this you need a stronger bound on $\frac{1}{\binom{n}{k}}$.

Tristan Liu
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As has been pointed out, the sum has a variable number of elements, so your original proof doesn't go through.

But ${n \choose 0} = {n \choose n} = 1$ and ${n \choose 1} = {n \choose n-1} = n$. So the sum is

$$2 + {2 \over n} + \sum_{k=2}^{n-2} {1 \over {n \choose k}}$$

Can you find a stronger bound on the terms of that sum?

Michael Lugo
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Just for your curiosity.

In year $1981$, Rocket proved the nice identity $$S_n=\sum_{k=0}^{n}\frac{1}{{n\choose k}}= \frac{n+1}{2^{n+1}}\sum_{k=1}^{n+1}\frac{2^k}{k}$$ Using special functions $$\sum_{k=1}^{n+1}\frac{2^k}{k}=-2^{n+2} \Phi (2,1,n+2)-i \pi$$ where $\Phi(.)$ is the Lerch transcendent function.

Claude Leibovici
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