Is This Piecewise Function Riemann Integrable?

[Lizzie11]

let f(x) = 1 when x in in [0,1)
f(x) = -1/2 when x is in [1,2)
f(x) = 1/3 when x is in [2, 3)

and so on, in othe words its the sequence (1/n)(-1)^n, whose series obviously converges to log 2. However is f(x) Riemann integrable and equal to this series?
If so, how to give an upper sum lower sum proof?, just choose a good partion?

thanks,

Lizzie

 

[fourier jr]

that function is riemann integrable but I'm not sure what it has to do with that sequence/series?

 

[Lizzie11]

Ok, but how to prove it?

It has to do with that sequence because on each interval [n, n+1) it is equal to (1/n+1)(-1)^n.

 

[HallsofIvy]

Ok, but how to prove it?
It has to do with that sequence because on each interval [n, n+1) it is equal to (1/n+1)(-1)^n.

Perhaps it would help if you clarified what your function is. In your original post you had
"let f(x) = 1 when x in in [0,1)
f(x) = -1/2 when x is in [1,2)
f(x) = 1/3 when x is in [2, 3)"

Which is a piecewise constant function that has trivially integrable. Yes, it is equal to (-1)^n/(n+1) on the interval [n,n+1) but what does that have to do with the series 1- 1/2+ 1/3+...

Ah, wait- you aren't asking whether the function is Riemann integrable- you are asking whether the definite integral from 0 to infinity exists! I think you've answered your own question. The improper integral $$\int_0^\infty f(x)dx$$ exists if and only if the limit $$lim_{A->0}\int_0^A f(x) dx$$ exists.

With f as you give it, $$\int_0^A f(x)dx= \Sigma_{i=1}^{N}\frac{(-1)^i}{i+1}$$ where N is the largest integer less than or equal to A and, as you say, that series converges to log 2.