can anyone show me how 


$$\displaystyle\frac{4}{R}\displaystyle\Pi\_{n=1}^{\infty} \left(\frac{1+R^{-4n}}{1+R^{-4n+2}}\right)^4= \frac{1}{R}\left(1+2 \sum\_{n=1}^ {\infty} \frac{1}{R^{2n(n+1)}}\right)^2\left(1+2 \sum\_{n=1}^{\infty}\frac{1}{R^{2n^{2}}}\right)^{-2} $$ 


The above identity is obtained by Komatu [page no, 58] (Y. Komatu, "A coefficient problem for function univalent in an annulus", Kodai math. Rep., 8, 1956, 49-70.)


Jacobi's triple product identity is
$$\sum\_{n=-\infty}^{\infty} x ^{n^{2}}y^{2n} = \prod\_{m=1}^\infty (1-x^{2m})(1+x^{2m-1}y^2)(1+x^{2m-1}y^{-2}) $$


substitute $ x = \frac{1}{R^2}$ , $y= \frac{1}{R}$ 


$$ 1+ 2\sum\_{n=1}^\infty \frac{1}{ R^{2n(n+1)}} = \prod\_{n=1}^\infty (1- R^{-4m})(1+R^{-4m})(1+R^{-4m+4}) $$ 
how $ \;\;\;
(1+R^{-4m})(1+R^{-4m+4}) = (1+R^{-4m})^2 $ 


If Jacobi's triple product identity is
$$\sum\_{n=-\infty}^{\infty} x ^{n^{2}}y^n = \prod\_{m=1}^\infty (1-x^{2m})(1+x^{2m}y)(1+x^{2m}y^{-1}) $$


substitute $ x = \frac{1}{R^2}$ , $y= \frac{1}{R^2}$ 


$$ 1+ 2\sum\_{n=1}^\infty \frac{1}{ R^{2n(n+1)}} = \prod\_{m=1}^\infty (1- R^{-4m})(1+R^{-4m-2})(1+R^{-4m+2}) $$ 


for second equation, subs $ x= \frac{1}{R^2}, y= 1 $


$$ 1+ 2\sum\_{n=1}^\infty \frac{1}{ R^{2n^{2}}} = \prod\_{m=1}^\infty (1- R^{-4m})(1+R^{-4m})^2 $$ 



Expression for infinite product
First of all, the original identity has an exponent of $4$ on the left hand side that you have missed. Moreover it is not correct as stated, by simply comparing constant terms of both sides. However it is just a simple typo probably, because the following is true
$$\frac{4}{R}\prod\_{n=1}^{\infty} \left(\frac{1+R^{-4n}}{1+R^{-4n+2}}\right)^4= \frac{1}{R}\left(2+2 \sum\_{n=1}^ {\infty} \frac{1}{R^{2n(n+1)}}\right)^2\left(1+2 \sum\_{n=1}^{\infty}\frac{1}{R^{2n^{2}}}\right)^{-2}.$$
To prove it you just have to use some specializations of [Jacobi's triple product identity](https://en.wikipedia.org/wiki/Jacobi_triple_product). Indeed it says
$$\sum\_{n=-\infty}^{\infty} x^{n^2}y^n=\prod\_{m=1}^{\infty}(1-x^{2m})(1+x^{2m}y)(1+x^{2m}y^{-1})$$
substituting $x=\frac{1}{R^2}$ and $y=\frac{1}{R^2}$ gives
$$2+2\sum\_{n=1}^{\infty} \frac{1}{R^{2n(n+1)}}=2\prod\_{m=1}^{\infty} \left(1-\frac{1}{R^{4m}}\right)\left(1+\frac{1}{R^{4m}}\right)^2$$
and substituting $x=\frac{1}{R^2}$ and $y=1$ gives
$$1+2\sum\_{n=1}^{\infty}\frac{1}{R^{2n^2}}=\prod\_{m=1}^{\infty}\left(1-\frac{1}{R^{4m}}\right)\left(1+\frac{1}{R^{4m-2}}\right)^2.$$
Komatu's identity follows by dividing the first equation by the second and squaring.


