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 an≥bn, an>bn when n is odd, and an=b2n−1+b2n.

Assuming that

∞∑n=1an

converges, then

∞∑n=1bn=∞∑n=1(b2n−1+b2n)=∞∑n=1an

also converges. However,

∞∑n=1(an−bn)>0

Since an≥bn 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*