Prove that limit n tends to infinity $1 + 2 \sum_{k=0}^n1/\binom{n}{k} = e^2$
I have not been able to proceed ..tried many things like ratio of nck and nc(k+1)...also opened it.!! Not able to slove.!!
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Does $\sum_{k=0}^n/\binom{n}{k}$ imply $\sum_{k=0}^n\dfrac1{\binom nk}$? – lab bhattacharjee May 01 '14 at 03:02
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Does $/\binom{n}{k}$ mean $1/\binom{n}{k} = \frac{k!(n-k)!}{n!}$? – Dilip Sarwate May 01 '14 at 03:02
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yes....!!! i meant that only.!! – maths lover May 01 '14 at 03:05
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I don't believe it is true: have you written it correctly? – May 01 '14 at 03:14
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ya.!!!i have writen it correctly..!! – maths lover May 01 '14 at 03:15
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1http://math.stackexchange.com/questions/151441/calculate-sums-of-inverses-of-binomial-coefficients – lab bhattacharjee May 01 '14 at 03:19
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Where did you find this problem? – Mhenni Benghorbal May 01 '14 at 04:09
1 Answers
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It's not true.
Binomial coefficients are smallest when $k$ takes on the smallest or largest values possible, and they grow very fast as $k$ moves towards the middle.
Correspondingly, the reciprocals show the opposite pattern. Most of the terms of the sum are completely negligible!
In fact, for $k \in [2, n-2]$, we always have
$$\binom{n}{k} \geq \frac{n(n-1)}{2} $$
and therefore
$$ 1 + \frac{1}{n} + \frac{1}{n} + 1 \leq \sum_{k=0}^n \frac{1}{\binom{n}{k}} \leq 1 + \frac{1}{n} + \frac{1}{n} + 1 + \sum_{k=2}^{n-2} \frac{2}{n(n-1)} $$
and therefore
$$ 2 + \frac{2}{n} \leq \sum_{k=0}^n \frac{1}{\binom{n}{k}} \leq 2 + \frac{2}{n} + (n-3) \cdot \frac{2}{n(n-1)} $$
from which we can see that all three terms must converge to $2$ as $n \to \infty$.