Trying to integrate a summation of a unit step function.

[Unassuming]

Homework Statement

Define I(x)= I( x - x_n ) =

{ 0 , when x < x_n
{ 1, when x >= x_n.

Let f be the monotone function on [0,1] defined by

f(x) = \sum_{n=1}^{\infty} \frac{1}{2^n} I ( x - x_n)

where x_n = \frac {n}{n+1} , n \in \mathbb{N}.

Find \int_0^1 f(x) dx.

Leave your answer in the form of an infinite series.

Homework Equations

The Attempt at a Solution

I know the theorem, if f is monotone on [0,1], then f is Riemann integrable on [0,1]. I am also familiar with what the graph of this function looks like.

My calculation of what is under the function is,

\sum_{n=1}^{\infty} \frac{1}{2^n} \frac{1}{n^2 +3n +2}.


I need some advice for completing the problem (assuming that sum is correct).
Thanks

 

[HallsofIvy]

Homework Statement

Define I(x)= I( x - x_n ) =

{ 0 , when x < x_n
{ 1, when x >= x_n.

Let f be the monotone function on [0,1] defined by

f(x) = \sum_{n=1}^{\infty} \frac{1}{2^n} I ( x - x_n)

where x_n = \frac {n}{n+1} , n \in \mathbb{N}.

Find \int_0^1 f(x) dx.

Leave your answer in the form of an infinite series.

Homework Equations

The Attempt at a Solution

I know the theorem, if f is monotone on [0,1], then f is Riemann integrable on [0,1]. I am also familiar with what the graph of this function looks like.

My calculation of what is under the function is,

\sum_{n=1}^{\infty} \frac{1}{2^n} \frac{1}{n^2 +3n +2}.

If you mean this to be f(x), shouldn't it be a function of x? If you mean it to be the integral, then aren't you done?

I need some advice for completing the problem (assuming that sum is correct).
Thanks