Question

Let M be the subset of the Natural numbers consisting of those integers n such that no digit of n in its decimal expression is 5. Prove that
∑n∈M 1/n converges.

Answer 1

it's divergent n power is smaller than 1

Answer 2

from the comparison test....

Answer 3

I know that \[\sum_{1}^{\infty} 1/n =\infty \] but for some reason with the subset it converges, and I dont know why

Answer 4

wow this is pretty interesting
a proof using 9 instead of 5 can be found here
http://www.cut-the-knot.org/arithmetic/algebra/HarmonicSeries.shtml

Answer 5

i can't imagine this proof is trivial, as it is contained in hardy and wright

Answer 6

Hum I will check it out thanks!