Let $p(n)$ be the integer partition. It is already known that for $0 < x < 1$, we have


$\sum\_{j=0}^{\infty} p(j)x^j = \prod\_{i=1}^{\infty}\dfrac{1}{1-x^{i}}$.


But what if the sum on the left side stops on some $k \geq 0$? That is, do we have any clue about what the sum $\sum\_{j=0}^{k} p(j)x^j$ is for each $k \geq 0$?


My specific interest is when $x = 1/q$ where $q$ is the size of a finite filed whose characteristic is larger than 2. If this is not a difficult problem, I think the answer should be out there but it is just that I cannot find it so far. My naive conjecture is $\sum\_{j=0}^{k} p(j)x^j \approx 1/(1-x)$ although I haven't thought very deeply about how close they are.




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If the above solution is not known, it will be equally as great if I can hear whether


$\sum\_{j=0}^{(n-2)m} p(j)/q^j \approx (1 + q^{-1} + q^{-2} + \cdots + q^{-n})^{m}$, where $q$ is as same as above, where $0 \leq m \leq q$ (with as much detail as possible).



Bound of Partial Sum of Partition Generating Function
The series $\sum p(j) x^j$ converges for $|x|<1$, so for $x = 1/q$ the difference between the infinite series and the partial sum drops exponentially with $k$. The difference is asymptotically smaller than $c^k$ for any $1/q \lt c \lt 1$. In other words, a good approximation to the truncated series is the infinite series itself, which does not depend on $k$.


Your approximation $(1 + q^{-1} + q^{-2} + ... + q^{-n})^m$ is not good for large $m$ and $q$. That is $1 + m/q + {m \choose 2}/q^2 + ...$ instead of $1 + 1/q + 2/q^2+ ...$ so it is only right in the constant term. 


If it helps, the asymptotic growth rate for [partitions](http://en.wikipedia.org/wiki/Partition_%28number_theory%29) is known: $p(j) \sim \frac {1}{4\sqrt 3 ~j}\exp (\pi \sqrt{2j/3}).$ For the above, you only need that the growth rate is subexponential. You can use this to get a slightly better approximation than the infinite series since you can estimate the first few missing terms.


