Help: Finding limit of power of summation

Hello, I have a question regarding limits.

I am trying to solve the following question $\lim_{n\to\infty}\left(\sum_{r=1}^n\frac1{n\choose r}\right)^n$

Now, I know how to solve infinite summations and $\lim_{x\to\infty}\left(1+\frac1x\right)^x$ , however, I don't know how to solve a combination of both, with a few more terms. The sum expands to $\frac1n+\dots+1$ . With only these (without the " $\dots$ " terms, of course) I can easily solve the limit, however, the extra terms add another layer of complexity.

How do I deal with the rest of the terms? The rest of the terms are higher powers of $n$ which should vanish (faster?) as $n\to\infty$ . This, probably, would make the answer $e$ . Am I right?

Easy Math Editor

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Comments

The answer is $e^2$ , not $e$ .

Hint: Arrange the terms of the sum in ascending order: $1, 1/n, 1/n, 2/(n(n-1)), 2/(n(n-1)), \ldots$ .

Hint 2: If $x\approx 0$ , then $1 + x + O(x^2) \approx 1 + x$ .

Hint 3: Apply $\displaystyle \lim_{x\to\infty} \left(1 + \frac1x\right)^x = e$ .

Pi Han Goh - 3 years, 1 month ago

Thanks. Apparently I needed to consider both the $1\over n$ s. Silly of me.

Sherlock Doyle - 3 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$