The geometric sum to infinity of a series with a ratio of r, and started with a, is calculated to be:

G(infinity) = a/(1-r)

 

But I am thinking of:

Each term in a series could be represented by:

a(0) = a; a(n) = ar^n,

so why can't we integrate a(n) with respective to dn, giving us (ar^n)/(ln®), but obviously this is the the expected answer. I'm interested in what makes such a difference. Thanks.

Briefly, the difference between [imath]\sum f(x)[/imath] and [imath]\int f(x)[/imath] is that you're not counting the values between the integers.

 

Have you done Riemann sums at all? The Riemann integral is the limit of the Riemann sums of a function as the partitions get finer, but in this case the partitions have a constant width of 1.

If you take a look at the attached image, you should see how the areas under the two graphs are not the same.

That's very clear, thanks.