# Analysis + convergence

1. Jan 17, 2006

### Pearce_09

$$f_n(x)= x^n/1 + x^n$$

does this series converge on the interval [0,1]
Say if x = 1 then the series is < some epslon , where epslon is > 0
but if x = 1 then the value for $$f_n(x)$$ is constant for all n
but does this still mean the series converges.. even tho the series doesnt get smaller for large values of n.....
if not then it doesnt converge?

Last edited: Jan 17, 2006
2. Jan 17, 2006

### fourier jr

3. Jan 17, 2006

### HallsofIvy

I assume you mean $f_n(x)= \frac{x^n}{1+ x^n}$ since otherwise (i.e. if you mean $f_n(x)= 2x^n$) I see no reason for the "1". Also It's not clear whether you mean a "sequence" rather than "series" (which is an infinite sum) since you say "even tho the series doesnt get smaller for large values of n". A sequence doesn't have to "get smaller", just get closer to (or be) a number (an infinites series does not converge it the sequence does not go to 0). fn(0)= 0 for all n and the "constant" sequence 0 certainly converges to 0 (and the infinite sum is also 0!). fn(1)= 1 for all n so the sequence "converges" to 1 but that means that the infinite series $\sum f_n(1)$ does not converge. For x any number between 0 and 1, we can divide both numerator and denominator by xn to get $\frac{1}{1+\frac{1}{x^n}}$. That sequence also converges to 0 so the series may converge. Since these are all positive values, we can apply the integral test: The series $\sum \frac{x^n}{1+ x^n}$ will converge if and only if the infinite integral $\int_1^\infty \frac{x^y}{1+ x^y} dy$ converges (for 0< x< 1). To integrate that, let u= 1+ xy. Then du= ln(x) xy so
$$\int_1^\infty \frac{x^y}{1+ x^y} dy= \frac{1}{ln x}\int_{1+x}^1\frac{du}{u}= -\frac{ln(1+x)}{ln(x)}$$
Okay, that exists (and is positive since x< 1), and so the series converges for all x< 1.
The sequence of functions converges to the discontinuous (at 0) function f(x)= 0 for 0<= x< 1, f(1)= 1.
The series converges on [0, 1).

Last edited by a moderator: Jan 17, 2006
4. Jan 17, 2006

### D H

Staff Emeritus
It's much easier to apply the ratio or comparison test.
Let $s_n = \sum_{n=0}^\infty f_n(x)$
Obviously sn diverges for x>=1 since fn converges to 1 for x=1 and diverges for x>1. Thus we are only concerned with abs(x)<1. Note that each term fn in sn is positive for all x>0 and that the terms alternate in sign for x<0.

Ratio test: Each term in sn is positive for all x>0.
sn is a convergent series if $f_{n+1}(x)/f_n(x) < 1$ for all sufficiently large f. The ratio converges to x for x<1, so sn is convergent for $x\in [0,1)$.

Comparison test: The series $\sum_{n=0}^\infty x^n$ converges for all x in (-1,1). For all x>0, each term $f_n(x) = \frac{x^n}{1+x^n}$ is smaller than $f_n(x) = x^n$. Thus sn is convergent for $x\in [0,1)$.

Alternating series test: The sequence $\frac{\lvert x^n\rvert}{1+x^n}$ is monotonically deceasing for $x\in(-1,0)$ and some sufficiently large n. Thus the alternating series \sum_{n=0}^\infty f_n(x), x\in(-1,0)[/itex] is convergent.

Last edited: Jan 17, 2006
5. Jan 17, 2006

### maverick6664

It looks easy....

If $$|x| < 1$$ it obviously converges to 0.

If $$x = 1$$ it is always $$1/2$$.

If $$x = -1$$ it doesn't converge, considering $$n$$ can take odd values.

If $$|x| > 1,$$ $$f_n(x)=x^n/(1+x^n) = 1/(1/x^n+1)$$, so it converges to 1.

Last edited: Jan 17, 2006