1. ## Series estimation

Let $A:=\left\{ n\in\mathbb{N}:\exists a,b\in\mathbb{N},\, a,b\geq2:\, n=a^{b}\right\}$. Prove that $$\underset{n\in A}{\sum}\frac{1}{n}\leq\frac{8}{9}.$$

2. ## Re: Series estimation

\begin{align*}\sum_{n \in A} \dfrac{1}{n} & = \sum_{a\ge 2}\left(\sum_{b\ge 2} \dfrac{1}{a^b}\right) \\ & = \sum_{a\ge 2}\left(\sum_{b\ge 0} \left(\dfrac{1}{a}\right)^b - 1 - \dfrac{1}{a}\right) \\ & = \sum_{a\ge 2}\left(\dfrac{1}{1-\tfrac{1}{a}} - 1 - \dfrac{1}{a}\right) \\ & = \sum_{a\ge 2} \left(\dfrac{1}{a-1} - \dfrac{1}{a}\right) \\ & = \left(1 - \dfrac{1}{2}\right) + \left(\dfrac{1}{2} - \dfrac{1}{3}\right) + \cdots\end{align*}

The $n$-th sum is:

$S_n = \sum_{a=2}^{n+1} \left(\dfrac{1}{a-1} - \dfrac{1}{a}\right) = 1 - \dfrac{1}{n+1}$

$\lim_{n \to \infty} S_n = 1$

So, I am pretty sure you cannot prove that sum is less than or equal to $\dfrac{8}{9}$.

3. ## Re: Series estimation

Originally Posted by SlipEternal
\begin{align*}\sum_{n \in A} \dfrac{1}{n} & = \sum_{a\ge 2}\left(\sum_{b\ge 2} \dfrac{1}{a^b}\right) \\ & = \sum_{a\ge 2}\left(\sum_{b\ge 0} \left(\dfrac{1}{a}\right)^b - 1 - \dfrac{1}{a}\right) \\ & = \sum_{a\ge 2}\left(\dfrac{1}{1-\tfrac{1}{a}} - 1 - \dfrac{1}{a}\right) \\ & = \sum_{a\ge 2} \left(\dfrac{1}{a-1} - \dfrac{1}{a}\right) \\ & = \left(1 - \dfrac{1}{2}\right) + \left(\dfrac{1}{2} - \dfrac{1}{3}\right) + \cdots\end{align*}

The $n$-th sum is:

$S_n = \sum_{a=2}^{n+1} \left(\dfrac{1}{a-1} - \dfrac{1}{a}\right) = 1 - \dfrac{1}{n+1}$

$\lim_{n \to \infty} S_n = 1$

So, I am pretty sure you cannot prove that sum is less than or equal to $\dfrac{8}{9}$.
Maybe I'm wrong but I think $$\underset{n\in A}{\sum}\frac{1}{n}<\underset{a,b\geq2}{\sum}a^{-b}.$$ For example in the second sums the number $16$ is added like $2^{4}$ and $4^{2}$.

4. ## Re: Series estimation

I see what you are saying...

Let's try to see when we add term more than once: For each $a\ge 2$, $a^{c\cdot d} = (a^c)^d$.

Some examples: $2^4 = 4^2 = 16, 2^6 = 4^3 = 8^2, 2^8 = 16^2, 2^{10} = 4^5 = 32^2$

So, $\sum_{n \in A} \dfrac{1}{n} \le \sum_{a\ge 2}\sum_{b\ge 2} a^{-b} - \sum_{a\ge 2}\sum_{b\ge 2} a^{-2b}$

$\sum_{a\ge 2}\sum_{b\ge 2} a^{-b} - \sum_{a\ge 2}\sum_{b\ge 2} a^{-2b} = \sum_{a\ge 2}\left(\dfrac{1}{a-1} - \dfrac{1}{a} - \dfrac{1}{a^2-1} + \dfrac{1}{a^2}\right)$

Now, the n-th partial sum is $S_n = 1 - \dfrac{1}{3} - \dfrac{1}{n+1} + \dfrac{1}{(n+1)^2}$ and $\lim_{n \to \infty} S_n = \dfrac{2}{3}$

This still isn't quite right, but it might give you some ideas. Now, I am not adding 1/4^4 at all.

5. ## Re: Series estimation

$\sum_{n \in A} \dfrac{1}{n} \stackrel{?}{=} \sum_{p\text{ is prime}}\sum_{b\ge 2} p^{-b} = \sum_{p\text{ is prime}}\left(\dfrac{1}{p-1} - \dfrac{1}{p}\right) = \left(1-\dfrac{1}{2}\right) + \left(\dfrac{1}{2} - \dfrac{1}{3}\right) + \left(\dfrac{1}{4} - \dfrac{1}{5}\right) + \cdots$

Then, this is equal less than or equal to $1-\dfrac{1}{3} + \dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{10} < \dfrac{8}{9}$

6. ## Re: Series estimation

No, this still misses some. For instance, powers of 6.

Let $P = \{n\in \Bbb{N} \mid n\ge 2, \forall b\in \Bbb{N}, b\ge 2, n^{1/b} \notin \Bbb{N} \}$

Then $\sum_{n \in A} \dfrac{1}{n} = \sum_{n \in P}\sum_{b\ge 2} n^{-b}$

I think that is true. Then $\sum_{n \in P}\sum_{b\ge 2}n^{-b} = \sum_{n\in P}\left(\dfrac{1}{n-1} - \dfrac{1}{n}\right) = \left(1-\dfrac{1}{2}\right) + \left(\dfrac{1}{2} - \dfrac{1}{3}\right) + \left(\dfrac{1}{4} - \dfrac{1}{5}\right) + \left(\dfrac{1}{5} - \dfrac{1}{6}\right) + \left(\dfrac{1}{6} - \dfrac{1}{7}\right) + \left(\dfrac{1}{9} - \dfrac{1}{10}\right) + \cdots$

That is no bigger than $1-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{9} < \dfrac{8}{9}$

7. ## Re: Series estimation

Originally Posted by SlipEternal
No, this still misses some. For instance, powers of 6.

Let $P = \{n\in \Bbb{N} \mid n\ge 2, \forall b\in \Bbb{N}, b\ge 2, n^{1/b} \notin \Bbb{N} \}$

Then $\sum_{n \in A} \dfrac{1}{n} = \sum_{n \in P}\sum_{b\ge 2} n^{-b}$

I think that is true. Then $\sum_{n \in P}\sum_{b\ge 2}n^{-b} = \sum_{n\in P}\left(\dfrac{1}{n-1} - \dfrac{1}{n}\right) = \left(1-\dfrac{1}{2}\right) + \left(\dfrac{1}{2} - \dfrac{1}{3}\right) + \left(\dfrac{1}{4} - \dfrac{1}{5}\right) + \left(\dfrac{1}{5} - \dfrac{1}{6}\right) + \left(\dfrac{1}{6} - \dfrac{1}{7}\right) + \left(\dfrac{1}{9} - \dfrac{1}{10}\right) + \cdots$

That is no bigger than $1-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{9} < \dfrac{8}{9}$
I don't understand how I can prove that $$\underset{n\in P}{\sum}\left(\frac{1}{n-1}-\frac{1}{n}\right)\leq\frac{8}{9}$$ you have added only few terms.

8. ## Re: Series estimation

Look at estimation of error.

$\sum_{n\in P}\left(\dfrac{1}{n-1} - \dfrac{1}{n}\right) = \left(1-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{7}\right)+\sum_{n\in P\setminus\{2,3,5,6,7\} }\left(\dfrac{1}{n-1} - \dfrac{1}{n}\right)$

Since $\sum_{n\in P\setminus\{2,3,5,6,7\} }\left(\dfrac{1}{n-1} - \dfrac{1}{n}\right)$ is an alternating sum (similar to a telescoping sum), it is no bigger than its first term, which is $\dfrac{1}{9}$. This part should not be difficult to prove.