Geometric series

Interesting, sort of

A sort of interesting limit of the first derivative is \(r\rightarrow1\). We investigate it using \(r=1+\epsilon\): \[\sum \limits_{k=0}^{n} a k =\lim_{\epsilon\rightarrow0} a \frac{n (1+\epsilon)^{n+1}-(n+1)(1+\epsilon)^n +1}{\epsilon^2}\]

Expanding the binomials to the quadratic term: \[\sum \limits_{k=0}^{n} a k =\lim_{\epsilon\rightarrow0} a \frac{n \left(1+(n+1)\epsilon+\dfrac{(n+1)n}{2}\epsilon^2\right)-(n+1)\left(1+n\epsilon+\dfrac{n(n-1)}{2}\epsilon^2\right) +1 +O(\epsilon^3)}{\epsilon^2}\]

This looks like it will blow up in the limit. But there are some convenient, but expected cancellations: \[\sum \limits_{k=0}^{n} a k =\lim_{\epsilon\rightarrow0} a \frac{n (n+1)n\epsilon^2-(n+1)n(n-1)\epsilon^2 +O(\epsilon^3)}{2\epsilon^2}\]

We are left with: \[\sum \limits_{k=0}^{n} a k =\lim_{\epsilon\rightarrow0} a \frac{n(n+1)}{2} (1+O(\epsilon))=a \frac{n(n+1)}{2}\]

This is the well-known formula (multiplied by a constant) for the sum of the arithmetic series, \[\sum\limits_{k=0}^n k=\frac{n(n+1)}{2}\]

By differentiating further, and taking appropriate limits, in principle one could derive sums of higher integer powers. We could also derive, using the same technique, the obvious \[\lim_{r \to 1}\sum\limits_{k=0}^{n}a r^{k+1}=\sum\limits_{k=0}^{n}a =(n+1)a\]

These sums are in fact dealt with better using Bernoulli polynomials.