How can I find sum $\sum_{n=k}^{\infty} \frac{\binom{n}{k}}{2^n}$? I tried differentiating binomial, but it didn't really help :(
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1Also: https://math.stackexchange.com/q/1941603/42969, https://math.stackexchange.com/q/1933175/42969 – Martin R Feb 26 '24 at 09:31
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I deleted my answer because this is a duplicate, and instead posted it as an answer to the original question. You’d asked why the probabilities sum to $1$. This is because the sum is over the probabilities for disjoint events, one of which will eventually occur at some point with probability $1$. The probability that you conduct an infinite sequence of Bernoulli trials with non-zero success probability and never get $k+1$ successes is $0$. – joriki Feb 26 '24 at 09:44
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@MartinR: You can add a second original to the duplication notice. I think it would be worthwhile to add https://math.stackexchange.com/questions/1941603 because it has more different derivations (I added mine there as well). (I can’t do it because I don’t have the golden calculus badge. :-) – joriki Feb 26 '24 at 09:48
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1@joriki: Done... – Martin R Feb 26 '24 at 09:51