Hi!
$$\lim_{n \to \infty }\int_{0}^{n}\frac{dx}{1+n^{2}\cos^{2}x }$$
I found to solution on the internet but I didn't understood it 100%.
First, it says that the function under integral has period $pi$.Why pi ? I know that cos function has period $2kpi$
Consequence: $\int_{0}^{k\pi...
Let $x_{0}=1$ and $x_{n+1}=(-1)^{n}(\frac{\pi }{2}-\arctan(\frac{1}{x_{n}}))$
I have the following options to choose from:
1. $x_n$ is unbounded
2. $x_n$ is increasing and the limit of $x_n$ is $1$
3. the limit of $x_n$ is $\pi/2$.
4. the limit of $x_n$ is $0$
My attempt:
I used...
Why the following limit doesn't exists ?
$$\lim_{x\rightarrow 0}xe^{-\frac{1}{x}}$$
I think it's because of $\frac{1}{x}$ which doesn't exists, right ?
I have the following sequence $(x_{n})$ , $x_{n}=1+\frac{1}{2^{2}}+...+\frac{1}{n^{2}}$ which has the limit $\frac{\pi ^{2}}{6}$.I need to calculate the limit of the sequence $(y_{n})$, $y_{n}=1+\frac{1}{3^{2}}+...+\frac{1}{(2n-1)^{2}}$
I don't know how to start.I think I need to solve the limit...
Let $0<b<a$ and $(x_{n})_{n\in \mathbb{N}}$ with $x_{0}=1, \ x_{1}=a+b$
$$x_{n+2}=(a+b)\cdot x_{n+1}-ab\cdot x_{n}$$
a) If $0<b<a$ and $L=\lim_{n\rightarrow \infty }\frac{x_{n+1}}{x_{n}}$ then $L= ?$
b) If $0<b<a<1$ and $L=\lim_{n\rightarrow \infty }\sum_{k=0}^{n}x_{k}$ then $L= ?$
I don't know...
Hi!
I have the following sequence $$(x_{n})_{n\geq 1}, \ x_{n}=ac+(a+ab)c^{2}+...+(a+ab+...+ab^{n})c^{n+1}$$
Also I know that $a,b,c\in \mathbb{R}$ and $|c|<1,\ b\neq 1, \ |bc|<1$
I need to find the limit of $x_{n}$.
My attempt is in the picture.The result should be $\frac{ac}{(1-bc)(1-c)}$
I...
$$a^{n}\leq a^{\frac{n}{1}}+a^{\frac{n}{2}}+...+a^{\frac{n}{n}}$$ this I can see is true, it's obvious
$$a^{n}\leq n\cdot a^{n}$$ like the first one, I can see it's true
$$a^{\frac{n}{1}}+a^{\frac{n}{2}}+...+a^{\frac{n}{n}}\leq n\cdot a^{n}$$ this one,I can't see "how it's true", it's not so...
Hello!
$$\lim_{n\rightarrow \infty }\frac{1}{n}ln(a^{\frac{n}{1}}+a^{\frac{n}{2}}+...+a^{\frac{n}{n}} ), \ a>1$$
I solved the limit by using the following inequality:
$$a^{n}\leq a^{\frac{n}{1}}+a^{\frac{n}{2}}+...+a^{\frac{n}{n}}\leq n\cdot a^{n}$$
After I applied a $ln$ and $1/n$ I got $lna$...