# how can I convert a sum series to an equivalent product series

There are many identities in which on one side we have a summation series and on the other side we have a product series. For e.g. the very famous rogers-ramanujan identity ∑_(n=0)^∞▒q^(n^2 )/〖(q;q)〗n =∏(n=1)^∞▒1/((1-q^(5n-1))(1-q^(5n-4))), here 〖(q;q)〗_n is known as Pochhammer symbols which is defined in inbuilt mathematica. Now I have learnt that it is possible to convert a sum series, as the L.H.S. of above e.g., to an equivalent product series using some mathematica code. can anyone plz help in making that mathematica code..

• Please give things properly formatted either in $\LaTeX$ or (preferred) in Mathematica. The jumble of symbols there is really not very helpful - it's in fact ambiguous. – Patrick Stevens Sep 7 '15 at 7:34
• Sir I want to convert a sum series (which is power series in q) to equivalent product series such that both have the same power series expansion. For e.g. we have such identity known as Roger-Ramanujan identitiy which is \begin{eqnarray}\label{c1} \sum_{\lambda=0}^{\infty}\frac{q^{\lambda^2}}{(q;q)_{\lambda}}=\prod_{\lambda=1}^{\infty}\frac{1}{(1-q^{5\lambda-1})(1-q^{5\lambda-4})},\nonumber \end{eqnarray} – Megha Goyal Sep 7 '15 at 8:20
• Now I want to convert \sum_{\lambda=0}^{\infty}\frac{q^{\lambda^2+2\lambda}}{(q;q)_{\lambda}} to an equivalent product series using mathematica... can you please help me in making some mathematica module to convert this sum series to equivalent product series. – Megha Goyal Sep 7 '15 at 8:24
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THIS IS NOT AN ANSWER BUT RATHER AN EXTENDED COMMENT

Mathematica can evaluate both the product and sum in closed form

f1[q_] = Product[1/((1 - q^(5 k - 1)) (1 - q^(5 k - 4))), {k, 1, Infinity}]

(* ((1 - 1/q^4)*(-1 + q))/
(q*QPochhammer[1/q^4, q^5]*
QPochhammer[1/q, q^5]) *)

Limit[f1[q], q -> 0]

(* 1 *)

f2[q_] = Sum[q^(k^2)/QPochhammer[q, q, k], {k, 0, Infinity}]

(* 1/(QPochhammer[q, q^5]*
QPochhammer[q^4, q^5]) *)

f2

(* 1 *)


While Mathematica does not recognize these as equal

Assuming[{0 < q < 1}, f1[q] == f2[q] // FullSimplify]

(* ((1 - 1/q^4)*(-1 + q))/
(q*QPochhammer[1/q^4, q^5]*
QPochhammer[1/q, q^5]) ==
1/(QPochhammer[q, q^5]*
QPochhammer[q^4, q^5]) *)


Mathematica can show that the series expansions are equal to any arbitrary degree

With[{n = 150}, (Series[f1[q], {q, 0, n}]) === (Series[f2[q], {q, 0, n}])]

(* True *)

With[{n = 150}, Series[f1[q] - f2[q], {q, 0, n}]]

(* SeriesData[q, 0, {}, 151, 151, 1] *)


And that the functions are numerically equal for any specific values of q

With[{delta = .000120},
And @@ Table[f1[q] == f2[q], {q, delta, 1 - delta, delta}]]

(* True *)
`
• thnx for your explaination.. but there is a code in mathematica which people use to convert a sum series to a product series. Can anybody plz help me. – Megha Goyal Sep 8 '15 at 2:13