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Aug 8, 2019 at 3:05 review Reopen votes
Aug 8, 2019 at 13:02
Jul 29, 2019 at 20:53 vote accept user2718
Jul 29, 2019 at 20:50 comment added Todd Trimble @Kostya_I Yes, thanks very much, but let's perhaps try to be generous as Matt F. suggests.
Jul 29, 2019 at 20:50 review Reopen votes
Jul 30, 2019 at 1:27
Jul 29, 2019 at 20:39 comment added user44143 @Kostya_I, the question uses the word "classical", which could be charitably interpreted here as "coming from a function satisfying a differential equation with rational coefficients". That may make the question more interesting.
Jul 29, 2019 at 20:33 comment added Steven Landsburg I failed to note condition that the coefficients are integers and voted to hastily to close. Apparently it's too late to retract the close vote, so voted to reopen.
Jul 29, 2019 at 20:31 comment added Iosif Pinelis @Kostya_I : I had the same idea, recorded in my answer, except it must be the largest, not the smallest. :-)
Jul 29, 2019 at 20:30 history closed Arturo Magidin
YCor
user44191
Kostya_I
Steven Landsburg
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Jul 29, 2019 at 20:28 comment added Kostya_I @ToddTrimble, isn't it obvious that we can cook up such a series for any positive number smaller than $1$? Just take $a_0:=0$ and inductively $a_i$ to be the largest positive integer such that the partial sum is still smaller than 1?
Jul 29, 2019 at 20:19 answer added Iosif Pinelis timeline score: 5
Jul 29, 2019 at 20:04 comment added Todd Trimble To people voting to close: is the answer so obvious? Let's discard the smart-aleck solution $a_0 = 1, a_i = 0$ for $i \geq 1$.
Jul 29, 2019 at 20:01 comment added user44143 The Taylor series for $\ln(1+x)$ has fractional coefficients, so it doesn't seem to help...but I don't see an obvious argument that settles this.
Jul 29, 2019 at 20:00 review Close votes
Jul 29, 2019 at 20:35
Jul 29, 2019 at 19:58 history edited user44143 CC BY-SA 4.0
corrected spelling, improved formatting
Jul 29, 2019 at 19:41 history asked user2718 CC BY-SA 4.0