# Study of the convergence (pointwis&uniform) of two series of functions

1. Dec 19, 2013

### Felafel

1. The problem statement, all variables and given/known data

study the pointwise and the uniform convergence of

$f_{n1}(x)=ln(1+x^{1/n}+n^{-1/x}$ with $x>0$ , $n \in |N^+}$ and $f_{n2}(x)=\frac{x}{n}e^{-n(x+n)^2}$ with $x \in \mathbb{R}$ , $n \in }|N^+}$
3. The attempt at a solution

1) first series: $f_{1n}$
studying the limit for n to infinity i found out it is ln2, so it converges pointwise to this value, but being the function increasing it doesn't have a maximum, and thus Weierstrass' criterion for uniform convergence doesn't yield.
however, for any compact in (0, infinity), say [a,b] with b>a it has a maximum in b. Thus, the function is also uniformly convergent for any compact in (0, infinity) but not in all $\mathbb{R}^+$.

2) second series: $f_{2n}$
studying the limit for n to infinity i found out it is 0, so it converges pointwise to this value.

am I missing some parts of the study/did I make any mistakes?

Last edited: Dec 19, 2013
2. Dec 19, 2013

### Felafel

sorry, i've made a bit of a mess, i'm trying to correct myself:
1) f_1n convergese pointwise to ln(2) as previously said, but I'm not studying the uniform convergence differently:
according to the definition, for n big enough i get:
$|f_n(x)-f(x)| < \epsilon$ $\forall \epsilon >0$, $\forall x in (0,\infty)$
which yields:
$|ln(1+x^{1/n}+n^{-1/x}-ln2| < \epsilon$ using the properties of logarithms i get:
$\frac{1+x^{1/n}+n^{-1/x}}{2}|<e^{\epsilon}$ 1 is the inferior of $e^{\epsilon}$ so can rewrite it in place of $e^{\epsilon}$ itself. then,
$\frac{1}{2}+\frac{x^{1/n}+n^{1/x}}{2}<1$
$\sqrt{x}+n^{-1/x}<1$ which is true $\forall x \leq \$

3. Dec 19, 2013

### haruspex

I guess you meant $^n\sqrt{x}+n^{-1/x}<1$, but I'm not sure what you intended here:
To prove uniform convergence this way, you want it true for all x > 0. It is not true for x ≥ 1.
Maybe it is not uniformly convergent. How would you prove that?

4. Dec 19, 2013

### Felafel

isn't it actually false even for x<1?
if i get x=1/2, for instance,
$(1/2)^{1/n}+n^{-2}$ goes to 1+0 if n is big.
so all the values x can have involve a contradiction with the definition of uniform convergence, as the $sup|f_n(x)-f(x)|$ isn't less than epsilon

5. Dec 19, 2013

### Felafel

also, if there aren't values of x that go on well with the definition of uniform convergence, can I say that the function converges uniformely in any compact subset of |R for the aforemetioned reasons (beginning of the thread)?

6. Dec 19, 2013

### Felafel

and, as far as the second series is concerned, i've just computed what follows:
$|\frac{x}{n} e^{-n(n+x)^2}|$ = $|\frac{x}{n} \frac{1}{e^{-n(n+x)^2}}| \leq |\frac{x}{n} \frac{1}{1+n(n+x)^2}|$ given $e^x \geq x+1$
$\leq |\frac{x}{n} \frac{1}{n(n+x)^2}|$ dividing num and denom by x i see it goes to 0, which means
$\sum||f_{n2}||_{\infty}=0$ and so weierstrass criterion for uniform convergence holds.

7. Dec 19, 2013

### haruspex

It tends to 1, but that doesn't prove it persistently exceeds 1.
But even if so, all you have shown here is that this method of proving uniform convergence does not work. It doesn't prove convergence is not uniform. How would you prove that?

8. Dec 20, 2013

### Felafel

you are probably right in saying convergence is not uniform, but ii really keep not seeing why my method doesn't work. i'll try to write things a bit differently. by the deginition of uniform convergence i have to prove there exists a n big enough such that:
$|f_n(x)-f(x)|< \epsilon$
$|ln(1+x^{1/n}+n^{-1/x}-ln(2)|\leq\epsilon$ $\forall x>0$
$|ln\frac{1+x^{1/n}+n^{-1/x}}{2}| \leq |ln\frac{1+1+\frac{1}{n}}{2}=|ln(1+\frac{1}{2n})|$ using the exp:
$|1+\frac{1}{2n}| \leq e^{\epsilon}$ i choose $\epsilon=\frac{1}{n^2}$ and see that for n big wnough this inequality holds.

if it's wrong again how could i start proving it doesn't converge uniformly?
also, the method i used for the second series in my last post was okay?

thank you very much for being so patient

9. Dec 20, 2013

### haruspex

I just noticed the title says 'series' not sequences. And you refer to Weierstrass' M-test, yes? That is indeed for series. But the functions in the f1n converge to ln(2), as you say, so their sum as a series cannot converge even pointwise.
So please clarify - are these sequences or series?
You can take the definition of uniform convergence and invert it logically.
The definition:
Its logical inversion:

Last edited: Dec 20, 2013
10. Dec 20, 2013

### Felafel

sorry again, they are sequences. yes, i was referring to weierstrass' m-test. can't I use it for sequences too?

11. Dec 20, 2013

### Felafel

i've tried to do as you suggested:
there exists an ε > 0 such that for every natural number N there exists x ∈ S and and n ≥ N with |fn(x) − f(x)| > ε

$|ln(1+x^{1/n}+n^{-1/x}-ln(2)| > \epsilon$
$|ln\frac{1+x^{1/n}+n^{-1/x}}{2}| > \epsilon$
if x=1
$|ln\frac{1+1+n^{-1}}{2}| > \epsilon \Rightarrow |ln(2+\frac{1}{n}|\geq \epsilon$
taking$\epsilon=ln2$ i should have proved the contraddiction. so it can't converge uniformly on all |R.

here's my attempt at proving for which subsets it converges uniformly:
i want to know for which values of x this inequality holds:
$|ln(1+x^{1/n}+n^{-1/x}-ln(2)| < \epsilon$
$|\frac{(1+x^{1/n}+n^{-1/x})}{2} < e^{\epsilon}$
$|\frac{(1+x^{1/n}+n^{-1/x})}{2} < \epsilon+1$
$x^{1/n}+n^{-1/x}<2 \epsilon +1$
$x^{1/n}<2 \epsilon +1$
$x < (2 \epsilon +1)^n$ for $\epsilon \to 0$ i get $x<1$
so it converges uniformly $\forall$ compact subset [a,b] where b<1.

Last edited: Dec 20, 2013
12. Dec 20, 2013

### haruspex

This logic is not going to work. The implications are backwards from what you need. You haven't proved |fn(x) âˆ’ f(x)| > Îµ, but that something rather larger > Îµ.
Algebraic error.
To make it easier, let's first look at the individual functions $x^{\frac 1n} -1$ and $n^{-\frac 1x}$. Each converges pointwise to 0. You need to get some idea of how the rate of converges depends on x. So, for each, consider how large n has to be to make the function < Îµ.

13. Dec 20, 2013

### Felafel

any further hint? i really can't make it :(, i had the following results but they seem absurd to me.
i tried to put:
$sup|f_n(x)-f(x)|=| sup(f_n(x)) - inf f(x)|= |sup (ln(1+x^(1/n)+n^(-1/x))| - ln2| = sup | ln(\frac{1+x^(1/n)+n^(-1/x))}{2}|$
now if x=1 i get $lim |ln(1/2+1/2+1/n)|=ln1=0$ so i have uniform convergence for this value. if $n<+\infty$ i wouldn't have u. convergence, though
if x=a s.t. a>1 i get $lim |ln(1+a^(1/n)+n^(-1/a))|=ln1=0$ same as above
if x=1/a s.t. a>1 i get $lim |ln(1+\frac{1}{a}^(1/n)+n^(-a))|=ln1=0$ same as above
so apparently i'd have uniform convergence for every x>0

14. Dec 20, 2013

### haruspex

You can't have 'uniform convergence' for a value of x. That doesn't mean anything.
Try rewriting $|x^{\frac 1n}-1| < \epsilon$ as a constraint on the value of n.

15. Dec 20, 2013

### Felafel

do you mea that i have to find n s.t. $1/n<log_x(1+\epsilon)$? And for $n>\frac{1}{log_x(1+\epsilon)$ it converges uniformly? From which passage did i get that expression?

16. Dec 20, 2013

### haruspex

Sort of. Because of the modulus operation we should treat x > 1 and x < 1 separately.
For x > 1, what you wrote is correct, but let's write it as n > ln(x)/ln(1+Îµ).
So far the logic chain has been reversible (have to be careful with that when dealing with inequalities). That is, the sequence converges uniformly if and only if for every Îµ > 0 we can find an N s.t. n > N implies n > ln(x)/ln(1+Îµ).
Well, is that true? Or is it the case that no matter what N we pick there is an x > 1 which will make ln(x)/ln(1+Îµ) > N?

17. Dec 21, 2013

### Felafel

doesn't it hold for any x?because the logarithmic function goes to infinity more slowly than n. Shoul I assume from that that it is uniformly convergent in all |R+?

18. Dec 21, 2013

### haruspex

You don't seem to have grasped what uniform convergence means.
For pointwise convergence, given an Îµ and an x, you find an N = N(Îµ, x). I.e. the choice of N is allowed to depend on x.
For uniform convergence, given an Îµ you have to find an N = N(Îµ) that will work for all x.
For the functions $x^{\frac 1n}$, uniform convergence will require us to pick an N >= ln(x)/ln(1+Îµ) for all x > 1. But that is clearly not possible, since ln(x) is not bounded above.

19. Dec 21, 2013

### Felafel

ok! Now it's clearer. So, convergence is not uniform in |R, but how do I find out if there are subsets of |R where the convergence is uniform?

20. Dec 21, 2013

### haruspex

Suppose we limit x to being in [1, a]. Now can you find N=N(Îµ) such that N > n(x)/ln(1+Îµ) for all x in [1, a]? Next, what about x < 1? After that, try $n^{-\frac 1x}$.
One thing we have to be careful about here. There may be some cancellation between $n^{-\frac 1x}$ and $x^{\frac 1n}$ such that even though neither by itself converges uniformly there is some magic by which the sum does. (I'm pretty sure that can't happen, but we need to prove it doesn't.)