The series ∑∞n=11n\sum_{n=1}^\infty\frac1n diverges!

We all know that the following harmonic series

n=11n=11+12+13+

diverges and grows very slowly!! I have seen many proofs of the result but recently found the following: S=11+12+13+14+15+16+ >12+12+14+14+16+16+=11+12+13+=S.
In this way we see that S>S.

Can we conclude from this that S is divergent??

Answer

The proof can be made a bit more rigorous by setting
an=1n:1,12,13,14,15,16,bn=12(n+1)/2:12,12,14,14,16,16,
Note that anbn, an>bn when n is odd, and an=b2n1+b2n.

Assuming that
n=1an
converges, then
n=1bn=n=1(b2n1+b2n)=n=1an
also converges. However,
n=1(anbn)>0
Since anbn and an>bn when n is odd.

Now, (3) says that
n=1bn=n=1an
and (4) says that
n=1bn<n=1an
These last two statements are contradictory, so the assumption that (2) converges must be false.

Attribution
Source : Link , Question Author : User8976 , Answer Author : robjohn

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