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Abel's summation formula states that for two functions $f$ and $g$, with $f$ differentiable, we have $$\sum_{k=1}^n f(k)g(k)=G(n)f(n)-\int_1^n G(x)f'(x)\; dx \tag{$*$}$$ where $G(n)=\sum_{k=1}^n g(k)$ (it is possible to generalize to other bounds, but let's stick with this for now). Now consider a sum in which the "weight" function $f$ depends not only on $k$ but also on $n$: $$\sum_{k=1}^n f(k,n)g(k).$$ Does there exist a similar formula?

For example, consider the basic example $f(a,b)=ab$. Then $$\sum_{k=1}^n f(k,n)g(k)=n\sum_{k=1}^n kg(k)=n^2G(n)-n\int_1^n G(x)\; dx$$ by $(*)$. If we use $(*)$ directly, by assuming that the derivative $f'(x)$ in the integral is $\frac d{dx} f(x,n)$, we obtain the same result. Does this always hold? In other words: can we extend Abel's summation formula to read $$\sum_{k=1}^n f(k,n)g(k)=G(n)f(n,n)-\int_1^n G(x)\frac d{dx}f(x,n)\; dx?$$

Avi
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    Of course: fix $n$ and use Abel for the functions $f_n$ and $g$, where $f_n(x)=f(x,n)$. – Did May 21 '15 at 10:13
  • @Did I wasn't sure whether I could fix $n$, since the formula depends on it and moreover now we're introducing $n$ into the derivative and integral, which worried me. Want to make that an answer so I can accept it? – Avi May 21 '15 at 10:16

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