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Let for each integer $n\geq 1$ the sum of divisors function $\sigma(n)$, defined as $\sum_{d\mid n}d$.

Then an integer is said to be perfect, or a perfect number, if $\sigma(n)=2n$.

For example, $n=6$ is perfect since $\sum_{d\mid 6}d=1+2+3+6=2\cdot 6$.

As a motivation of the following claim, I've included two statements in an appendix.

Claim 3. If $n$ is an odd perfect number, then the following condition $$2 \left( \sum_{m=1}^{\sigma(2n)-1}m \right) \left( \sum_{k=0}^\infty\frac{1}{ \left( \sigma(2n) \right)^k } \right)= \left( \sigma(2n)\right)^2 \tag{1}$$ holds.

Question. Prove, or refute finding a counterexample, that if an integer $N>1$ satisifies $(1)$ then is an odd perfect number. Many thanks.

Appendix:

This is a genetic motivation of what were my previous claims related with previous one. You can write an easy proof as iff statements:

Claim 1. An integer $n>1$ is perfect if and only if satisfies $$2 \left( 1+2+3+\ldots+(n-1) \right)\sum_{k=0} ^\infty n^{-k}=\frac{n\sigma(n)}{2}.$$

And finally, (the same again)

Claim 2. Let $n>1$ an integer. Then $n$ is a perfect number if and only if $$\frac{1}{n} \left( \sum_{m=1}^{\sigma(n)-1}m \right)\sum_{k=0} ^\infty \left( \sigma(n) \right) ^{-k}=\sigma(n).$$

Remark. I've written these claims with LHS as slightly differents, to avoid misinterpretation. But the corresponding RHS were written intentionally in an attempt to get some interpretation (that I didn't get).

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    $$ \sum_{m=1}^{\sigma(n)-1}m=\frac{\sigma(n)}{2}(\sigma(n)-1) $$ – 2017-02-07
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    $$ \sum_{k=0}^\infty\frac{1}{\left(\sigma(n)\right)^k}=\frac{\sigma(n)}{\sigma(n)-1} $$ – 2017-02-07
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    @HazemOrabi many thanks for your attention, and if want to add more comments in next future explaining your ideas. – 2017-02-07
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    All users are welcome if want to criticize the question and the claims that I've deduced, I am saying about if the identities are/aren't interestings. Sometimes I try get statements about perfect numbers but I haven't avanced in the problem. – 2017-02-07

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$$ \sum_{m=1}^{M-1}m=\frac{M(M-1)}{2} \quad\&\quad \sum_{k=0}^{\infty}\frac{1}{M^k}=\frac{M}{M-1} \quad\colon\,M\in\mathbb{N}^{+} \\[4mm] n\,\in\,{perfect} \quad\buildrel\color{red}{\text{if and only if}}\over\iff\quad \sigma(n)=2n \quad\colon\,n\in\mathbb{N}^{+} $$
$$ \begin{align} &\, C1 =2\,\left(\sum_{m=1}^{n-1}m\right)\,\left(\sum_{k=0} ^{\infty}\frac{1}{n^k}\right) =2\,\frac{n(n-1)}{2}\,\frac{n}{n-1} =n^2 \\[2mm] &\, C1 = n^2 \color{red}{=\frac{n\,\sigma(n)}{2}} \quad\iff\quad \sigma(n)=2n \quad\iff\quad n\,\in\,{perfect} \\[8mm] &\, C2 =\frac{1}{n}\,\left(\sum_{m=1}^{\sigma(n)-1}m\right)\,\left(\sum_{k=0} ^{\infty}\frac{1}{\left(\sigma(n)\right)^k}\right) =\frac{1}{n}\,\frac{\sigma(n)\left(\sigma(n)-1\right)}{2}\,\frac{\sigma(n)}{\sigma(n)-1} =\frac{\left(\sigma(n)\right)^2}{2\,n} \\[2mm] &\, C2 = \frac{\left(\sigma(n)\right)^2}{2\,n} \color{red}{=\sigma(n)} \quad\iff\quad \sigma(n)=2n \quad\iff\quad n\,\in\,{perfect} \\[8mm] &\, C3 =2\,\left(\sum_{m=1}^{\sigma(2n)-1}m\right)\,\left(\sum_{k=0} ^{\infty}\frac{1}{\left(\sigma(2n)\right)^k}\right) =2\,\frac{\sigma(2n)\left(\sigma(2n)-1\right)}{2}\,\frac{\sigma(2n)}{\sigma(2n)-1} =\left(\sigma(2n)\right)^2 \\[2mm] &\, C3 = \left(\sigma(2n)\right)^2 \color{red}{=\left(\sigma(2n)\right)^2} \quad\iff\quad n\in\mathbb{N}^{+} \quad\Longleftarrow\quad n\,\in\,{odd\,\,perfect} \end{align} $$ Hence, although claim 3 is true as stated, it is better to notice that it holds for all positive integers including odd perfects. It is only a matter of simplifying LHS to RHS.

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    Many thanks one more time for your maths in this site. Thanks now from your proof I see my mistake, that is that Claim 3 is satisfied by all integers $n>1$. – 2017-02-08
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    @user243301: Correct. Thanks for U 2. – 2017-02-08