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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.

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 (with as much detail as possible) where $0 \leq m \leq q$.

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.

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).

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.

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 (with as much detail as possible).

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.

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 (with as much detail as possible) where $0 \leq m \leq q$.

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 $\sum_{j=0}^{k} p(j)x^j \approx x/(x-1)$ although I haven't thought very deeply about how close they are.

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 (with as much detail as possible).

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.

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 (with as much detail as possible).