# How to Solve a “Sum Inverse” [closed]

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.

## closed as unclear what you're asking by Théophile, Clayton, achille hui, Crostul, Markus ScheuerApr 9 at 21:11

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• Welcome to MSE. Do you have a question? I have voted to close for now because it isn't clear what you're asking. If you don't have a question, you might consider writing this as a blog post, for example. – Théophile Apr 9 at 19:39
• All of you've done is forced the elements of a partition of a set to be integer points. – Clayton Apr 9 at 19:58

Or if you want a shorter proof, if $$\sum_{k=1}^n a_k=b_n$$ then $$a_n=\sum_{k=1}^n a_k-\sum_{k=1}^{n-1}a_k=b_n-b_{n-1}$$.