Before I start, I want to say that the kinds of problems like (solve for $x$: $1=\sum_{n=1}^x(\ln(n))$) is not what I'm talking about. I'm talking about if you have a certain function like $\ln(x)$ and want to find what sums to that function. In other words, it's the opposite of finding a partial sum (for $\ln(x)$, it's $\ln(\frac{x}{x-1})$.
If you just want the answer straightforward, here it is: the inverse sum of a certain function $f(x)$ is $f(x)-f(x-1)$.
EDIT: I removed the proof because it was too confusing. Please look at J.G.'s answer for the proof.
Or if you want a shorter proof, if $\sum_{k=1}^n a_k=b_n$ then $a_n=\sum_{k=1}^n a_k-\sum_{k=1}^{n-1}a_k=b_n-b_{n-1}$.