Summing up an Arithmetic Progression via Integration?

[varadgautam]

Why doesn't the integration of the general term of an A.P. give its sum? Integration sums up finctions, so if I integrate the general term function of an A.P., I should get its sum.
Like
2,4,6,8,...
T=2+(n-1)2=2n
\int T dn=n^2 ..(1)

Sum=S=(n/2)(4+(n-1)2)=(n/2)(2+2n)=n+(n^2) ..(2)

Why aren't these two equal?

 

[micromass]

I see no reason at all why the integral should equal the sum. The integral doesn't sum integers, it calculates area.

That said, we do have the following inequality (this does not hold in general!):

\sum_{k=0}^{n-1}{f(k)}\leq \int_0^n{f(x)dx}\leq\sum_{k=1}^n{f(k)}

This inequality is the best you can do, I fear...

 

[chiro]

I see no reason at all why the integral should equal the sum. The integral doesn't sum integers, it calculates area.

That said, we do have the following inequality (this does not hold in general!):

\sum_{k=0}^{n-1}{f(k)}\leq \int_0^n{f(x)dx}\leq\sum_{k=1}^n{f(k)}

This inequality is the best you can do, I fear...

If you are dealing with a polynomial expression, you can use what are called Bernoulli polynomials.

If the expression is not a simple one (as in some finite polynomial expression), then the inequality is a good bet, unless there are some tighter constraints for the specific expression.

 

[nickalh]

Clarification:
Haven't you dropped a sign?
On the left hand integral, after integrating, I see
-ln|1 - x|

On the next or final line, the leading negative disappears.