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Let $f_{n}:\mathbb{R}^{2} \to\mathbb{R}$ be defined by $$f_{n}(x,y)=\frac{1}{n^{2}} \cdot \frac{x}{1+x+y}$$ Does $\sum_{n=1}^{\infty} f_{n}$ converge uniformly?

For testing the converges of $\sum f_{n}$ Weierstrass test is the only thing I know of. So here is what I have done.

We know that $x < 1+x+y$ and so $\frac{x}{1+x+y} < 1$ Now by comparison we have $$\sum f_{n} < \sum \frac{1}{n^{2}}$$ hence $\sum f_{n}$ converges uniformy. Am I correct with my argument. Can anyone tell me where it converges to. I think $\sum f_{n} \to 0$. But not sure.

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What happens when $x+y+1$ is close to $0$? (Or is that supposed to be $x^2+y^2+1$?) –  Yoni Rozenshein Mar 12 '13 at 8:14
    
@YoniRozenshein yeah i fogot abt that case. so it might nt cnverge uniformly –  Kangaroooo Mar 12 '13 at 8:24
    
Also, your inequality is not correct $1 \not< 1 + 1 + (-3)$. –  mrf Mar 12 '13 at 8:27
    
@MRF: THAT 1LY wrks fr +ve numbers –  Kangaroooo Mar 12 '13 at 8:28
    
But you include negative numbers in your domain ("$f_n : \Bbb{R}^2 \to \Bbb{R}$") –  Yoni Rozenshein Mar 12 '13 at 8:28

1 Answer 1

Calculating the limit (pointwise, in points where everything is defined) is easy, since the dependence on $n$ is only a multiplicative constant...

$$\sum_{n=1}^{\infty} f_n(x,y) = \sum_{n=1}^{\infty} \frac 1 {n^2} \cdot \frac {x} {1+x+y} = \frac {x} {1+x+y} \cdot \sum_{n=1}^{\infty} \frac 1 {n^2} = \frac {x} {1+x+y} \cdot \frac {\pi^2} 6$$

Regarding uniform convergence - even if you fix certain problems with your question, like the problematic line $1+x+y=0$, it does not globally converge uniformly. To illustrate why, we'll pick a simpler example:

Let's say $g_n : \Bbb{R} \to \Bbb{R}$ is given by $g_n(x) = \frac {x} {n}$. Obviously, $\lim_{n\to\infty} g_n(x) = 0$ pointwise. But does the sequence converge uniformly?

For uniform convergence to hold, we need the rate of decay to not depend on $x$. However, for any $\epsilon > 0$ and $n$, we can find some $x$ (maybe very big) such that $g_n(x) \ge \epsilon$. So this does not converge uniformly globally in $\Bbb{R}$. The sum in your question behaves similarly.

However, it does converge locally uniformly. This means if we pick a compact (closed and bounded) set $K$ (let's say this set does not intersect the line $x+y+1=0$), then the sum converges uniformly in $K$.

The proof of this is easy: Let $K$ be a compact set not interesecting $1+x+y=0$. The function $\frac x {1+x+y}$ is continuous in $K$ and therefore (from Weierstrass's theorem) bounded in $K$, let's say $\left| \frac x {1+x+y}\right| \le M_K$. Then we have for all $(x,y) \in K$,

$$\left| f_n(x,y) \right| \le M_K \cdot \frac 1 {n^2}$$

implying (by the Weierstrass M-test) uniform convergence in $K$, and therefore locally uniform convergence in $\Bbb{R}^2 \setminus \{1+x+y=0\}$.

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