Infinitely Print Zeno's Dichotomy Paradox (1/(2^n))
05AB1E, 10 9 bytes
Saved 1 byte thanks to Erik the Outgolfer
[No…+1/J?
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Explanation
[ # loop over N infinitely [0 ...]
No # calculate 2^N
…+1/J # join with the string "+1/"
? # print without newline
Python 2, 30 bytes
-5 thanks to Erik the Outgolfer
i=1
while 1:print i,'+1/';i*=2
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APL (Dyalog Unicode), 15 bytes
More fun if ⎕FR
(Floating-point Representation) is 1287
(128 bit decimal) and ⎕PP
(Print Precision) is 34.
{∇2×⊃⎕←⍵'+1/'}1
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{
…}1
apply the following function on the number 1:
⎕←⍵'+1/'
print the argument and the the string
⊃
pick the first one (i.e. the argument)
2×
double that
∇
tail call recursion on that (optimised, so it can be infinitely repeated)