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Jan 31, 2018 at 17:52 comment added Hans I suppose your question and answer generalize to when $\frac1{2^n}$ is replaced by $p^kq^{n-k}$ for $p\in(0,1)$?
Jan 31, 2018 at 17:44 comment added Hans Wow, I just happened upon your question. This is exactly a generalization of my question on both mathoverflow.net/q/291772/32660 and math.stackexchange.com/q/2625685/64809. Would you care to add your opinion on at least the mathoverflow.net site?
Apr 13, 2017 at 12:19 history edited CommunityBot
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Feb 21, 2011 at 6:15 history edited Mike Spivey
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Feb 19, 2011 at 0:59 vote accept Mike Spivey
Feb 17, 2011 at 23:26 vote accept Mike Spivey
Feb 17, 2011 at 23:30
Feb 17, 2011 at 23:17 answer added Mike Spivey timeline score: 5
Nov 22, 2010 at 1:14 comment added Mike Spivey @Gerry Myerson: Thanks. I am familiar with binomial transforms, but unfortunately nothing that I know about them has helped me thus far with this question.
Nov 21, 2010 at 23:29 comment added Gerry Myerson There is something called the "binomial transform" of a sequence, which is not exactly what you've got but close enough that maybe some of what's known about it could rub off.
Nov 21, 2010 at 21:19 comment added Mike Spivey @Gerry Myerson: If we use $\log n = H_n − \gamma +O(\frac{1}{n})$ (and taking $\log 0 = 0$ or just starting the summation at $k=1$) we get that the limit with $f(n)= \log n$ is $\ln 2$ once again. So I guess that proves that if $f$ does have logarithmic growth, then the limit will be finite and nonzero. So now we're just left with the question about functions with other kinds of sublinear growth.
Nov 21, 2010 at 0:28 comment added Gerry Myerson Can you evaluate the limit in the case $f(n)=\log n$?
Nov 20, 2010 at 23:10 answer added Aaron Meyerowitz timeline score: 1
Nov 20, 2010 at 21:46 history asked Mike Spivey CC BY-SA 2.5