Uniform Convergence

[jessebu]

The question is to prove that the series (sum from n=1 to ?) of x/(n(1+n+x)) converges uniformly on R.

I was trying to use the Weierstrass M-test, but I couldn't find a uniformly convergent function that was greater that this series for all n, x.

 

[tutor_joel]

Are you sure this is convergent? If you do some algebra and solve for a 1 in the numerator, you get this:

1/((n(n+1)/x + n))

Now, as you can see, with large values of x, the fraction goes to 1/n , a divergent series

however, if x is between zero and 1 there is a different story. What do you know about x? can it be anything? Just because the weiersterstrass test has the condition M(x) < M(n) for the given values of x

 

[jessebu]

We aren't given a range for x, but I understand what you are saying. Is there any way to manipulate the equation to show that it is less than 1/n^2? That's what I was trying to do, but I couldn't quite get there.

 

[galactus]

If we expand, we get:

\(\displaystyle \frac{x}{n(1+n+x)}=\frac{1}{n}-\left(\frac{1}{x+n+1}+\frac{1}{nx+n^{2}+n}\right)\)

 

[daon]

Are you sure this is convergent? If you do some algebra and solve for a 1 in the numerator, you get this:

1/((n(n+1)/x + n))

Now, as you can see, with large values of x, the fraction goes to 1/n , a divergent series

however, if x is between zero and 1 there is a different story. What do you know about x? can it be anything? Just because the weiersterstrass test has the condition M(x) < M(n) for the given values of x

x is fixed...