Find limit for sequence $(\frac n{n+1})_{n=1}^{\infty}.$

Need help in getting limit of sequence. This question is taken from section 3.2.1 of Chapter 3 of the CRM series book by MAA: Exploratory Examples for Real Analysis, By Joanne E. Snow, Kirk E. Weller.

I request to vet attempt; as well help to find the solution.

It is not solvable as L’ Hopital’s rule applies for $ \frac 0 0 $ or $ \frac {\infty}{\infty}$ cases alone, & the form is $ \frac 1{\infty}$ .

Q.3. Identify limit of Seq. 3 using calculus. $ $ (\frac n{n+1})_{n=1}^{\infty}$ $

Let, $ f(n) = (\frac n{n+1})_{n=1}^{\infty}$ , need find $ f'(n)\approx\frac{\Delta y}{\Delta n}= \lim_{\Delta n \to0 } (\frac{f(n+\Delta n)-f(n)}{(n+\Delta n)-n})_{n=1}^{\infty}$

$ = \lim_{h\to0 }(\frac{f(n+h)-f(n)}{h})_{n=1}^{\infty}$

$ =\lim_{h\to0 }(\frac{\frac {n+h}{n+h+1}-\frac n{n+1}}{h})_{n=1}^{\infty}$

$ =\lim_{h\to0 }(\frac{\frac {(n+h)(n+1) -(n)(n+h+1)}{(n+h+1)(n+1)}}{h})_{n=1}^{\infty}$

$ =\lim_{h\to0 }(\frac{\frac {(n^2+n+nh+h) -(n^2+nh+n)}{(n^2+n+nh+h+n+1)}}{h})_{n=1}^{\infty}$

$ =\lim_{h\to0 }(\frac{\frac {h}{(n^2+nh+2n+1)}}{h})_{n=1}^{\infty}$ $ =(\frac {1}{n^2+2n+1})_{n=1}^{\infty}$

What will be $\lim\limits_{x \rightarrow 0^+} \left ( 2 \sin {\frac {1} {x}} + \sqrt x \sin {\frac {1} {x}} \right )^x.$

Evaluate

$ $ \lim\limits_{x \rightarrow 0^+} \left ( 2 \sin {\frac {1} {x}} + \sqrt x \sin {\frac {1} {x}} \right )^x.$ $

I tried by taking log but it wouldn’t work because there are infinitely many points in any neighbourhood of $ 0$ where $ \ln \left ( \sin {\frac {1} {x}} \right )$ doesn’t exist. How to overcome this situation?

Any help will be highly appreciated.Thank you very much for your valuable time.

show this inequality to $\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3} $

Let $ a,b$ and $ c$ be positive real numbers. Prove that $ $ \sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3} $ $

This problem is from Iran 3rd round-2017-Algebra final exam-P3,Now I can’t find this inequality have solve it,maybe it seem can use integral to solve it?

Solving $u_{\epsilon \eta} = \frac{ u_\epsilon – u_eta}{4 (\epsilon – \eta) } .$

I’m trying to solve the following PDE

$ $ y^2 u_{xx} – u_{yy} = 0.$ $

I’ve found its canonical form as $ $ u_{\epsilon \eta} = \frac{ u_\epsilon – u_eta}{4 (\epsilon – \eta) } .$ $

However, I’m having trouble in solving this PDE. I’ve tried solving with mathematica, and that gave me

$ $ u = \frac{ x + y^3}{ 3} ,$ $ but, I need to know how to obtain such a result from the canonical form of the given PDE.

Estimate $\frac 1 {a+b} + \frac 1 {(a+b)(a+2b)} + \cdots + \frac 1 {(a+b)\cdots(a+kb)} + \cdots$

Suppose $ a>1,b>0$ are real numbers. Consider the summation of the infinite series: $ $ S=\frac 1 {a+b} + \frac 1 {(a+b)(a+2b)} + \cdots + \frac 1 {(a+b)\cdots(a+kb)} + \cdots$ $ How can I give a tight estimation on the summation? Apparently, one can get the upper bound: $ $ S\le \sum_{k=1} ^\infty \frac 1 {(a+b)^k}=\frac 1 {a+b-1}$ $ But it is not tight enough. For example, fix $ a\rightarrow 1$ , and $ b=0.001$ , then $ S=38.969939$ , it seems that $ S=O(\sqrt{1/b})$ . Another example: $ a=1$ , and $ b=0.00001$ ,$ S=395.039235$ .

On the determinant $\det[(\frac{i^2+dj^2}p)]_{0\le i,j\le(p-1)/2}$ with $(\frac dp)=-1$

Let $ p$ be an odd prime. For $ d\in\mathbb Z$ we define $ $ T(d,p):=\det\left[\left(\frac{i^2+dj^2}p\right)\right]_{0\le i,j\le(p-1)/2},$ $ where $ (\frac{\cdot}p)$ is the Legendre symbol. By (1.17) of my paper arXiv:1308.2900, if$ (\frac dp)=-1$ then $ (\frac{T(d,p)}p)=1$ .

Suppose that $ p\equiv3\pmod4$ . Then $ T(d,p)=T(-1,p)$ for any $ d\in\mathbb Z$ with $ (\frac dp)=-1$ . As $ T(-1,p)$ is a skew-symmetric determiant of even order, $ T(-1,p)$ is always a square.

In the case $ p\equiv1\pmod4$ , if $ d$ and $ d’$ are both quadratic nonresidues modulo $ p$ , then we clearly have $ T(d,p)=\pm T(d’,p)$ .

I have the following conjecture which seems quite challenging.

Conjecture. Let $ p\equiv1\pmod4$ be a prime and write $ p=x^2+4y^2$ with $ x$ and $ y$ positive integers. Then, for any integer $ d\in\mathbb Z$ with $ (\frac dp)=-1$ , there is a positive integer $ t(p)$ (not depending on $ d$ ) such that $ $ |T(d,p)|=2^{(p-1)/2}t(p)^2y.$ $

Via Mathematica, I find that \begin{gather}t(5)=1,\ t(13)=3,\ t(17)=4,\ t(29)=91,\ t(37)=81,\ t(41)=180, \t(53)=1703,\ t(61)=87120,\ t(73)=16104096,\ t(89)=3947892146, \ t(97)=19299520512,\ t(101)=885623936875,\ t(109)=36548185365.\end{gather}

Your comments are welcome!