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 Scheuer Apr 9 at 21:11

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  • 2
    $\begingroup$ 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. $\endgroup$ – Théophile Apr 9 at 19:39
  • $\begingroup$ All of you've done is forced the elements of a partition of a set to be integer points. $\endgroup$ – Clayton Apr 9 at 19:58

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


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