Tag: converges

Determine if $f_n(x)=\frac{1-x}{1+x^n}$ converges uniformly on $[0,2]$

I want to determine the uniform convergence of the family of functions $ f_n:[0,2]\to\mathbb{R}$ defined by $ $ f_n(x)=\frac{1-x}{1+x^n}. $ $

My attempt

  • Limit of $ f_n$

Let $ f$ be the limit of $ f_n$ , then $ $ f(x)=\begin{cases} 1-x,& 0\le x <1\ 0,&1\le x\le 2 \end{cases}. $ $ $ f$ is continuous on $ [0,1]$ , but not differentiable at $ x=1$ .

  • Inequalities about $ |f_n(x)-f(x)|$

We know that $ $ |f_n(x)-f(x)| = \begin{cases} \frac{(1-x)x^n}{1+x^n},&0\le x<1\ \frac{x-1}{1+x^n},&1\le x\le 2 \end{cases}. $ $ If $ 0\le x<1$ , $ $ |f_n(x)-f(x)|=\frac{(1-x)x^n}{1+x^n} \le (1-x)x^n, $ $ and $ (1-x)x^n$ has the maximum $ \frac{1}{n}\left(1-\frac{1}{n+1}\right)^{n+1}$ at $ x=\frac{n}{n+1}$ . Also, if $ 1\le x\le 2$ , then $ $ |f_n(x)-f(x)|=\frac{x-1}{1+x^n}\le \frac{x-1}{x^n} $ $ and $ \frac{x-1}{x^n}$ has the maximum $ \frac{n-2}{(n-1)^2}\left(1-\frac{1}{n-1}\right)^{n-1}$ at $ x=\frac{n}{n-1}$ . Thus we obtain \begin{align} |f_n(x)-f(x)| &\le \max\left\{\frac{1}{n}\left(1-\frac{1}{n+1}\right)^{n+1},\frac{n-2}{(n-1)^2}\left(1-\frac{1}{n-1}\right)^{n-1}\right\}\ &\le \frac{1}{n-1} \end{align} for sufficiently large $ n$ .

  • Conclusion

For $ \epsilon > 0$ , there is $ N\in\mathbb{N}$ such that $ \frac{1}{N}<\epsilon$ by Archimedian property. Then, for any $ n\in\mathbb{N}$ and $ x\in [0,2]$ , if $ n > N$ , then $ $ |f_n(x)-f(x)|\le \frac{1}{n-1}\le \frac{1}{N} <\epsilon. $ $ Therefore $ f_n$ uniformly converges on $ [0,2]$ .

“Solution” $ f_n$ does not converge uniformly, since $ f_n$ is differentiable but $ f$ is not differentiable at $ x=1$ .

However, I think the solution is wrong. For example, $ f_n(x)=\frac{\sin nx}{\sqrt{n}}$ converges uniformly on $ \mathbb{R}$ , whereas $ f_n'(x)=\sqrt{n}\cos nx$ does not pointwisely converge.

Question Is my attempt correct? If so, is there an easier method to solve the problem?

How can I prove that $\sum_n \left(\frac{n+1}{n^2+1}\right)^3$ converges?

Clearly $ \sum_n \left(\frac{n+1}{n^2+1}\right)^3$ converges, but I am having an embarrassingly hard time proving it.

I tried the Cauchy Root test, leading me to prove that $ \limsup \left(\frac{n+1}{n^2+1} \right)^\frac{3}{n}=1$ , implying that the root test cannot be used. Since the ratio test is equivalent to the root test, I cannot use that either.

Any suggestions of directions I can take?

Prove that this sequence converges to the golden ratio…

Question: Consider the sequence $ \{x_n\}$ defined by $ x_{n + 1} = 1 + 1/x_n$ for all $ n \in \mathbb{N}$ and with $ x_1 = 1$ .

  1. Prove that $ 0 < x_{2n + 2} \leq x_{2n}$ and $ x_{2n + 1} \geq x_{2n – 1} > 0$ .
  2. Prove that the subsequences $ \{x_{2n}\}$ and $ \{x_{2n – 1}\}$ both converge to the same limit.
  3. Prove that the sequence converges to $ \frac{1 + \sqrt{5}}{2}$ .

Answer:

  1. I was able to prove this part by mathematical induction.
  2. This is where I am getting stuck. My initial idea was to show that the sequence $ \{x_{2n} – x_{2n – 1}\} \to 0$ as $ n \to \infty$ , but this didn’t seem to get me anywhere. Here is my attempt.

Proof: Fix $ \epsilon > 0$ . Consider the following

$ $ |x_{2n} – x_{2n – 1}| = \bigg|\frac{1}{x_{2n – 1}} – \frac {1}{x_{2n – 2}}\bigg| \leq \bigg|\frac{1}{x_{2n – 1}}\bigg| + \bigg|\frac{1}{x_{2n – 2}}\bigg|.$ $

Now consider the sequence $ \{1/x_{2n – 1}\}$ . Since $ \{x_{2n – 1}\}$ is a positive increasing sequence, we have that $ 1/x_{2n – 1} < 1$ for all $ n \in \mathbb{N}$ . Thus, the sequence $ \{1/x_{2n – 1}\}$ converges by the monotone convergence theorem to some limit $ \alpha$ . This is where I was getting stuck. Any help would be nice.

  1. I saw that this question was linked here: Prove that the sequence $ (a_n)$ defined by $ a_0 = 1$ , $ a_{n+1} = 1 + \frac 1{a_n}$ is convergent in $ \mathbb{R}$ , but I was wondering if someone could elaborate more.

Proof that a sequence converges.

Let $ (a_n)_{n=1}^{\infty}$ be a sequence, let $ \; 0 < q < 1$ such that $ \;\forall n \in \mathbb{N}$ $ \;n>2$

$ \left | a_{n+1} -a_n \right | < q*\left | a_{n} -a_{n-1}\right |$

prove that $ (a_n)_{n=1}^{\infty}$ converges.

Well, I thought about proving it this way:

$ \left | a_{n+1} -a_n \right | < q*\left | a_{n} -a_{n-1}\right |<\left | a_{n} -a_{n-1}\right | $

therefore $ \left | a_{n+1} -a_n \right | <\left | a_{n} -a_{n-1}\right | \;$ foreach $ n>2$ and that obviously shows us that the sequence converges because the gaps between each 2 indexes reduces as $ n$ grows.

I have a feeling however that this is not a formal way of proving this, it seems to easy.

Cauchy sequence with $\{ x_n : n \in \mathbb{N} \}$ not closed converges

Suppose $ (x_n)_{n \in \mathbb{N}}$ is a Cauchy sequence and $ A = \{ x_n : n \in \mathbb{N} \}$ not closed. Show that there exists $ x \in X$ such that $ x_n \longrightarrow x$ .

Since $ A$ is not closed: $ $ \forall y\in A , \exists \varepsilon > 0 s.t. S(y,\varepsilon) \subset A $ $ and since $ x_n$ is Cauchy: $ $ \forall \varepsilon > 0 ,\exists n_0 \in \mathbb{N} s.t. \forall n,m \geq n_0 : \rho (x_n,x_m) < \varepsilon $ $ But I can’t see a way to connect these two to prove convergence. Some starting hints would be greatly appreciated.

Geometric distribution of a Yule process multiplied by $e^{-\lambda s}$ converges to exponential distribution with parameter 1

If Y is a Yule process with rate $ \lambda$ than I know that the generating function is $ $ F(x,t) = \frac{xe^{-\lambda t}}{1-x+xe^{-\lambda t}}$ $ .

Now I want to show that $ $ \lim_{s\to \infty}e^{-\lambda s}Y(s)=\epsilon$ $ where $ \epsilon$ has a exponential distribution with mean 1.

If I define $ Z(s):=e^{-\lambda s}Y(s)$ than the generating function should be $ $ F(x,s) = \frac{xe^{-\lambda s e^{-\lambda s}}}{1-x+xe^{-\lambda s e^{-\lambda s}}}$ $ But as $ s\to \infty$ I get $ $ \frac{x}{1-x+x}=x \neq \frac{1}{1-x}$ $ because $ $ \lim_{s \to \infty} e^{-\lambda s e^{-\lambda s}} = 1$ $ .

Where is my mistake?