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A divergent series …. or maybe not?

The number n ranges over all possible powers with both the base and the exponent positive integers greater than n, assuming each such value only once. Prove that:

    \[\sum_{n} \frac{1}{n-1}=1\]

Let us denote by \mathcal{M} the set of positive integers greater than 1 that are not perfect powers ( i.e are not of the form a^p , where a is a positive integer and p \geq 2 ).  Since the terms of the series are positive , we can freely permute them. Thus,

    \begin{align*} \sum_{n} \frac{1}{n-1} &= \sum_{m \in \mathcal{M}} \sum_{k=2}^{\infty} \frac{1}{m^k-1} \\ &= \sum_{m \in \mathcal{M}} \sum_{k=2}^{\infty} \sum_{j=1}^{\infty} \frac{1}{m^{kj}}\\ &=\sum_{m \in \mathcal{M}} \sum_{j=1}^{\infty} \sum_{k=2}^{\infty} \frac{1}{m^{kj}} \\ &= \sum_{m \in \mathcal{M}} \sum_{j=1}^{\infty} \frac{1}{m^j \left ( m^j-1 \right )} \\ &= \sum_{n=2}^{\infty} \frac{1}{n\left ( n-1 \right )} \\ &= \sum_{n=2}^{\infty} \left ( \frac{1}{n-1} - \frac{1}{n} \right )\\ &= 1 \end{align*}


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