So it's here http://en.wikipedia.org/wiki/Cauchy_condensation_test My question is this: is the value of the base 2 in 2^k an arbitrary value? Or is there something special about 2? Can we just use something like e^k instead?
Notice that the function f in [itex]\sum_{n=0}^{\infty} 2^{n}f(2^{n})[/itex] need not be well-defined for all arguments in the real numbers, but only for the natural numbers (including the zero). What you have is a positive monotone decreasing sequence (which is denoted here by [itex]f(n)[/itex] but could just as well be written as [itex]a_n[/itex]). Now consider the sum (infinite series): the idea now is to form summation blocks with length [itex]2^n[/itex] and find a useful estimate for each block. Every block incorporates [itex]2^n[/itex] summands and the biggest summand in each block is the first one, as the sequence is monotone decreasing. Try writing this down and see what you end up with. If you can, check the converse.