Let $A:=\left\{ n\in\mathbb{N}:\exists a,b\in\mathbb{N},\, a,b\geq2:\, n=a^{b}\right\}$. Prove that $$\underset{n\in A}{\sum}\frac{1}{n}\leq\frac{8}{9}.$$
I see what you are saying...
Let's try to see when we add term more than once: For each , .
Some examples:
So,
Now, the n-th partial sum is and
This still isn't quite right, but it might give you some ideas. Now, I am not adding 1/4^4 at all.
Look at estimation of error.
Since is an alternating sum (similar to a telescoping sum), it is no bigger than its first term, which is . This part should not be difficult to prove.