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The below identity I have found experimentally.

Question. Is this true? If so, may you provide a "slick" (or any) proof. $$6\sum_{k=1}^{\infty}\frac{k^2q^k}{(1-q^k)^2}+12\left(\sum_{k=1}^{\infty}\frac{kq^k}{1-q^k}\right)^2=\sum_{k=1}^{\infty}\frac{(5k^3+k)q^k}{1-q^k}.$$

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    $\begingroup$ How on earth did this make HNQ? (Of course, there's nothing at all wrong with it as a question; it's just rather... unexciting to be "hot".) $\endgroup$ – David Richerby Sep 17 '18 at 14:31
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Follow the comments of Lucia and note that $$\sum_{n\ge 1}\frac{n^2q^n}{(1-q^n)^2}=q\frac{\,d}{\,dq}\sum_{n\ge 1}\frac{nq^n}{1-q^n}.$$ I believe the identity actually is the well known $$q\frac{\,d}{\,dq}L=\frac{L^2-M}{12},$$ where $$L=1-24\sum_{n\ge 1}\frac{nq^n}{1-q^n}\;\mbox{ and}\; M=1+240\sum_{n\ge 1}\frac{n^3q^n}{1-q^n}.$$ You can find it in here https://en.wikipedia.org/wiki/Eisenstein_series.

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$$ \sum_{k=1}^\infty \frac{k^2 q^k}{(1-q^k)^2} = \sum_{n=1}^\infty \sigma(n) n q^n$$ $$ \sum_{k=1}^\infty \frac{k q^k}{1-q^k} = \sum_{n=1}^\infty \sigma(n) q^n$$ $$ \left(\sum_{k=1}^\infty \frac{k q^k}{1-q^k}\right)^2 = \sum_{n=1}^\infty \sum_{m=1}^{n-1} \sigma(m) \sigma(n-m) q^n $$ $$ \sum_{k=1}^\infty \frac{k^3 q^k}{1-q^k} = \sum_{n=1}^\infty \sigma_3(n) q^n $$ so your identity is saying

$$ 6 n \sigma(n) + 12 \sum_{m=1}^{n-1} \sigma(m)\sigma(n-m) = 5 \sigma_3(n) + \sigma(n)$$

Hmm, surely that's got to be known. Adding number-theory to the tags.

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    $\begingroup$ Such identities are obtained by multiplying Eisenstein series and decomposing the product in terms of modular forms. See Ramanujan's paper On certain arithmetical functions which includes this identity, Table IV line 1. ramanujan.sirinudi.org/Volumes/published/ram18.pdf $\endgroup$ – Lucia Sep 17 '18 at 6:45
  • $\begingroup$ Good comment, as always. $\endgroup$ – T. Amdeberhan Sep 17 '18 at 13:34

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