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Using central limit theorem , evaluate

$\lim_{n\to \infty}\sum_{j=0}^{n}{j+n-1 \choose j}(\frac{1}{2^{n+j}})$,

I multiplied and divided the series by $1/2$ , And made it look like a binomial distribution ,but they are not i.i.d., which is why I cannot apply CLT.

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  • $\begingroup$ I think you are confusing distributions and random variables $\endgroup$ – seanv507 Oct 31 '17 at 16:46
  • $\begingroup$ Then how shall i proceed? $\endgroup$ – DRPR Oct 31 '17 at 16:47
  • $\begingroup$ For a given $n$, interpret the sum as the probability of an event related to a series of coin-flipping experiments. See en.wikipedia.org/wiki/Negative_binomial_distribution for some suggestive formulas. The limit will then be the chance of a particular event associated with a standard Normal variable. $\endgroup$ – whuber Oct 31 '17 at 16:55
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Some elaborations on whuber's hint:

Suppose you flip a fair coin repeatedly. What is the probability that your $n$th heads appears on the $(n+j)$th flip? Knowing this, can you write your sum as the probability of some event?

Your sum is the probability of getting [at least] $n$ heads in $2n$ flips.

Then, use what you know the Central Limit Theorem says about the distribution of heads in a number of coin flips (binomial distribution), as the number of coin flips tends to infinity.

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