## How to solve recursion with two seperate converges rates

What is the correct way to solve the following recursion: $$T(n)=T(\lceil\frac{n}{2}\rceil) + T(n-2)$$

Or basically any recursion that has two parts which converge in a different rate.

I’m trying to get big $$O$$ approximation, as tight as possible, but I couldn’t figure it out with any "traditional" approach.

## 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?

## Sequence of integrable functions on [a,b] converges pointwise but not uniformly?

I’m currently taking Real Analysis and there was an example in the textbook showing that the sequence of integrable functions fn(x) converges pointwise. I’m a bit confused as to why that is and also why does it not converge uniformly? Here is the example in textbook!

Thank you!

## Sequence of integrable functions on [a,b] converges pointwise but not uniformly?

I’m currently taking Real Analysis and there was an example in the textbook showing that the sequence of integrable functions fn(x) converges pointwise. I’m a bit confused as to why that is and also why does it not converge uniformly? Here is the example in textbook!

Thank you!

## 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 $$\frac{1 + \sqrt{5}}{2}$$.

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.

## Find the sum of an infinite series if it converges

How would I go about finding the sum of the following infinite series?

$$\sum\limits_{n=1}^{\infty} \frac{1}{1+2+3+4+..+n}$$

I’m assuming a limit of some sort. I apologize if this is trivial, I am not that great at math.

## 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?