Prove that the following sequences of functions converges to zero

Question

Let$\ f_n (x)=n^2x(1-x)^n$ I need to prove that$\ f_n→0$ in the interval$\ [0,1]$.

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Let$\ f_n(x) = nx^n$ prove that$\ f_n→0$ in the interval$\ [0,1)$.

For both of these sequences I tried the following:

By taking the function$\ f(x)=0$ we can see that

$$\lim_{n\rightarrow\infty}f_n(x) = 0$$

for both of the sequences, but I don't know if this is the correct way of solving both problems and I have my doubts if this even means that$\ f_n→0$, I'm thinking that what I did before actually implies that$\ f_n→f$.

Help would be greatly appreciated.

Answer 1

The usual thing for this type of problem is to look at the derivative to see how the functions behave.

If $f_n(x) =n^2x(1-x)^n $, then $f_n(0) = f_n(1) = 0$.

Also,

$f_n'(x) =n^2((1-x)^n-xn(1-x)^{n-1}) =n^2(1-x)^{n-1}(1-x-xn)\\ =n^2(1-x)^{n-1}(1-x(n+1)) $

so $f_n'(x) = 0$ at $x_n = \frac1{n+1}$.

At this point, $f_n(x_n) =n^2\frac1{n+1}(1-\frac1{n+1})^n \sim n/e $ since, as $n \to \infty$, $\frac{n}{n+1} \to 1$ and $(1-1/n)^n \to 1/e$.

Therefore $f_n$ can be arbitrarily large for large enough $n$ and does not uniformly converge to zero.

However, for any given $x \in (0,1)$,

$\begin{array}\\ n^2(1-x)^n &=\exp(2\ln(n)+n\ln(1-x))\\ &=\exp(2\ln(n)-n(x+x^2/2+...))\\ &\lt\exp(2\ln(n)-nx))\\ &\to 0 \qquad\text{since }2\ln(n)-nx \to -\infty \\ \end{array} $

since $\frac{\ln(n)}{n} \to 0$ as $n \to \infty$. Therefore $\lim_{n \to \infty}f_n(x) = 0$.

if $n^2$ is replaced by $n$, then the max value is about $1/e$, so $f_n$ again does not converge uniformly to z

However, if the $n^2$ is replaced by a constant (usually $1$), then the max value is about $1/(ne)$ so the sequence uniformly converges to zero.

Answer 2

One has $f_n(0) = f_n(1) = 0$, and if $0<x< 1$ is fixed, then $0<1-x<1$ and hence, for any fixed $k\in \mathbb{N}$ (not only $2$ as in your example) $$ (1) \qquad n^k(1-x)^n \to 0 \text{ as } n \to \infty, $$ and hence $f_n(x) \to 0$.

To see $(1)$, one can (for instance) consider the logarithm of the sequence, namely $$ \log n^k(1-x)^n = k \log n + n \log (1-x) \to -\infty, $$ where the latter is due to the fact that the coefficient of $n$ is a fixed strictly negative number, and $n$ grows much faster than the logarithmic function.

The convergence, however, is not uniform, as was pointed in the answer by @martycohen: the maximum of $f_n$ is at $\frac{1}{n+1}$ where the function stays away from $0$.

Answer 3

For the first one, we have $f_n(0) = f_n(1) = 0$, so certainly the sequence converges to zero at $x=0$ and $x=1$. For $0<x<1$, observe that $$\frac{f_{n+1}(x)}{f_n(x)} = \frac{(n+1)^2}{n^2}(1-x),$$ so $$\lim_{n \to \infty}\frac{f_{n+1}(x)}{f_n(x)} = 1-x$$ Since $0 < 1-x < 1$, the ratio test implies that the series $\sum_{n=1}^{\infty}f_n(x)$ converges, and therefore the sequence $f_n(x)$ must converge to zero.

You can prove the second one quite similarly.

Answer 4

$f_n(x)=nx^2(1-x)^n$, $x \in [0,1].$

1) $f_n(x)=0$ for $x=0$ or $x=1$.

2) Let $x \in (0,1)$.

Set $y:=1-x$ , where $0<y<1$.

$y^n= \dfrac {1}{z^n}$, where $z: =1/y >1$.

Write $z=1+t$, where $t >0$.

Then:

$f_n(x)=n^2x \dfrac{1}{(1+t)^n} \lt $

$n^2x \dfrac{1}{1+..n(n-1)(n-2)t^3/3!+..}$

$\lt n^2x\dfrac{3!}{(n-3)^3t^3}$.

Hence:

$0<f_n(x)< \dfrac{xn^2}{n^3(1-3/n)^3t^3}$.

Let $n>6$ then $(1-3/n) >1/2$:

$0<f_n(x) < \dfrac{2^3x}{nt^3}=(\dfrac{2^3x}{t^3})\dfrac{1}{n}$.

Squeeze!