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Math Help - Series estimation

  1. #1
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    Series estimation

    Let $A:=\left\{ n\in\mathbb{N}:\exists a,b\in\mathbb{N},\, a,b\geq2:\, n=a^{b}\right\}$. Prove that $$\underset{n\in A}{\sum}\frac{1}{n}\leq\frac{8}{9}.$$
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  2. #2
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    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}.
    Last edited by SlipEternal; June 9th 2014 at 10:17 AM.
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  3. #3
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    Re: Series estimation

    Quote Originally Posted by SlipEternal View Post
    \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 $4^{2}$.
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  4. #4
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    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.
    Last edited by SlipEternal; June 9th 2014 at 12:02 PM. Reason: \dfrac{1}{a^a} is added in \sum_{a\ge 2}\sum_{b\ge 2} \dfrac{1}{a^b} only once for each a, not twice.
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  5. #5
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    Re: Series estimation

    Oh, how about this:

    \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}
    Last edited by SlipEternal; June 9th 2014 at 12:39 PM.
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  6. #6
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    Re: Series estimation

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

    How about this:

    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}
    Last edited by SlipEternal; June 9th 2014 at 01:50 PM.
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  7. #7
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    Re: Series estimation

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

    How about this:

    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.
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  8. #8
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    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.
    Last edited by SlipEternal; June 10th 2014 at 05:24 PM.
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