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)