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I was trying to evaluate the limit of the sum $$\lim_{n\to\infty} \sum_{i=0}^n \frac{1}{_nC_i}$$ My first idea was, since the number of terms grows but the size of each term shrinks as $n \to \infty$, to turn this into an integral, because an integral is basically the same thing. But I'm not quite sure how to turn this into an integral, and I don't even know if it can be done.

Can it be done this way? If so, can I have a hint as to how?

If not, how could I find this limit? Just by observing its values, it seems to approach $2$, but again, I can't figure out how to show that.

Franklin Pezzuti Dyer
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2 Answers2

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The terms corresponding to $i=0$ and $i=n$ contribute $1$ each to the sum. The terms for $i=1$ and $i=n-1$ contribute $1/n$ each to the sum. The other terms contribute at most $1/\binom n 2$ to the sum; there are $n-3$ of them. So your sum differs from $2$ by $O(1/n)$.

kimchi lover
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$$\binom{n}{k}=\frac{n!}{k!(n-k)!}=\frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma(n-k+1)}=\frac{1}{n+1}\frac{1}{\frac{\Gamma(k+1)\Gamma(n-k+1)}{\Gamma(n+2)}}=\frac{1}{(n+1)B(k+1,n-k+1)}=\frac{1}{(n+1)\int_{0}^{1} x^k(1-x)^{n-k}dx}$$

$$\sum_{k=0}^{n}\frac{1}{\binom{n}{k}}=(n+1)\int_{0}^{1}\sum_{k=0}^{n}x^k(1-x)^{n-k}dx$$

$$\sum_{k=0}^{n}x^k(1-x)^{n-k}=(1-x)^n\sum_{k=0}^{n}x^k(1-x)^{-k}=\frac{x^{n+1}-(1-x)^{n+1}}{2x-1}$$

MAN-MADE
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