a question about the sum of $ e^{ikx}$

Question

I have a simple question. Let $ S = \sum\limits_{k = 1}^n {e^{ikx} } $ using the typical trick , we also have $ S\left( {e^{ix} - 1} \right) = e^{i\left( {n + 1} \right)x} - e^{ix} $ and if $ \left( {e^{ix} - 1} \right) \ne 0 $ we divide by it and we have: $ S=\left( {1 - e^{inx} } \right)\frac{{e^{ix} }} {{1 - e^{ix} }} $ And I see that this limit if n goes to infinity does not exist, but Wolfram|Alpha says that it converges. The problem clearly is in the factor $ \left( {1 - e^{inx} } \right) $ it goes to 0 :S? or something

Answer 1

Assuming $x$ is real, Wolfram is wrong.

Answer 2

When evaluating the sum, Wolfram Alpha makes the necessary assumptions so that the series converges. For example Sum[x^k, {k, 0, Infinity}] evaluates to $1/(1-x)$. It is not mentioned that this is only valid for $|x| < 1$, but that doesn't mean $1 + 2 + 2^2 \ldots = -1$

Similarly, Sum[Exp[i x k], {k, 1, Infinity}] evaluates to $-1 + 1/(1 - e^{ix})$ under the assumption that $|e^{ix}| < 1$ or $\mbox{Imag}(x)>0$. If you want Mathematica to output the conditions necessary for convergence, you need the GenerateConditions flag, like this:

Sum[Exp[i x k], {k, 1, Infinity}, GenerateConditions -> True]

Unfortunately this doesn't work in Alpha. In Mathematica it evaluates to: $$ \mbox{ConditionalExpression} \left[-\frac{e^{i x}}{-1 + e^{i x}}, e^{\mbox{Re}[i x]} < 1 \ \ \&\& \ \ e^{i x} \ne 1 \right] $$ Since you didn't specify that $x$ must be real, it is assumed to be possibly complex. Alternatively, one can specify that $x$ is real with a statement like this:

Assuming[ Element[x, Reals], Sum[Exp[I k x], {k, 1, Infinity}] ]

As expected, this gives the error Sum::div: Sum does not converge..

So, why doesn't Mathematica give the convergence conditions by default? Probably just because it results in a simpler output. (If one is really interested in finding the analytic continuation of a function, the convergence conditions don't matter.)