Derivative of a sequence Just a quick question, can you derive sequences like any normal funtion ?
For example, if $a_n=\frac{1}{n}$ then $\frac{d}{dn}(a_n)=-\frac{1}{n^2}$, is this ok ?
 A: The comparable thing is to use the Finite Difference operator.  The forward difference is defined as $\Delta_1[a_n]=a_{n+1}-a_n=\frac 1{n+1}-\frac 1n=\frac {-1}{n^2+n}$, not too different from $\frac {-1}{n^2}$.  It can be useful, particularly with series that are polynomials.  Each application of the difference operator reduces the degree by $1$, so if you keep applying it you can find the degree and then the whole polynomial.
A: This depends on your sequence. For the example you give in particular, the answer is no.
The derivative of $f$ at $c$ (for $c$ finite) is defined as the following limit (if it exists): $$ f'(c) =  \lim_{x \to c}\frac{f(x)-f(c)}{x-c}.$$
By $\lim_{x \to c}$ we mean that we need to be able to take $x$ as close as we want to $c$ in order for this to even make sense.
The issue with your sequence $a_n = \frac{1}{n^2}$ is that $a_n$ is technically speaking a function defined on the natural numbers: we might write it as $a_n = a(n)$. Limiting processes do not make sense in the natural numbers: if $c$ is not a natural number, then there is no way to get "infinitely close" to $c$ in the natural numbers.
On the other hand, we could instead have a sequence of functions defined on the real numbers: take for example $$f_n(x) = x^n.$$
We can differentiate the terms of this sequence with respect to $x$, and get $f_n'(x) = nx^{n-1}.$
A: There is finite calculus which is exactly what you are describing. Here is a paper which is very well written, easy to read introduction to finite calculus which starts with the basics and develops the rules for discrete differentiation and integration. Discrete integration is equivalent to summation so the ultimate goal of this paper is application of finite calculus "for solving nasty sums".
