How find that $\left(\frac{x}{1-x^2}+\frac{3x^3}{1-x^6}+\frac{5x^5}{1-x^{10}}+\frac{7x^7}{1-x^{14}}+\cdots\right)^2=\sum_{i=0}^{\infty}a_{i}x^i$ let
$$\left(\dfrac{x}{1-x^2}+\dfrac{3x^3}{1-x^6}+\dfrac{5x^5}{1-x^{10}}+\dfrac{7x^7}{1-x^{14}}+\cdots\right)^2=\sum_{i=0}^{\infty}a_{i}x^i$$
How find the $a_{2^n}=?$
my idea:let
$$\dfrac{nx^n}{1-x^{2n}}=nx^n(1+x^{2n}+x^{4n}+\cdots+x^{2kn}+\cdots)=n\sum_{i=0}^{\infty}x^{(2k+1)n}$$
Thank you everyone

Let $\chi_2(n)$ be the Dirichlet character modulo $2$
Define $\sigma'(n)=\sum_{d\mid n}d\chi_2(d)$
$$\sum_{n=1}^\infty\sigma'(n)x^n=\sum_{n=1}^\infty \frac{\chi_2(n)nx^n}{1-x^{2n}}=\sum_{n \text{ odd}}\frac{nx^n}{1-x^{2n}}$$
$$(\sum_{n \text{ odd}}\frac{nx^n}{1-x^{2n}})^2=\sum_{n=2}^\infty\sum_{k=1}^{n-1}\sigma'(n)\sigma'(n-k)x^n=\sum_{n=0}^\infty a_nx^n$$
With $a_0=0$, $a_1=0$, and for $n>1$
$$a_n=\sum_{k=1}^{n-1}\sigma'(n-k)\sigma'(k)$$
If your looking for an elementary evaluation of this, and your not familiar with the theory of elliptic functions you could try reading this guys blog, https://math.stackexchange.com/users/72031/paramanand-singh which uses elementary properties of trigonometric functions and the manipulations of several power series to prove convolution identities, if this is too much for you, you could try looking up basic combinatoral identities which can be used to give 'elementary' evaluations of divisor sum convolutions. For example Skoruppa's combinatorial identity: $$\sum_{\substack{ax+by \\ (a,b,x,y)\in \mathbb{N^{4}}}}h(a,b)-h(a,-b)=\sum_{d\mid n} (\frac{n}{d}h(d,0)-\sum_{j=0}^{n-1}h(d,j))$$ For a function satisfying $h(y,y-x)=h(x,y)$
For example, note that $$(x-y)^2+x^2+y^2$$
Satisfies this constraint and upon substitution gives the nice identity, $$\sum_{k=1}^{n-1}\sigma(k)\sigma(n-k)=\frac{5\sigma_3(n)}{12}-\frac{(6n-1)\sigma(n)}{12}$$
