Linked Questions

3
votes
2answers
163 views

algebrainc proof for $\alpha\beta \leq \frac{\alpha^p}{p} + \frac{\beta^q}{q}$ for conjugate exponents p and q [duplicate]

I found a very informal proof (geometric one, for by taking $\alpha$ as $\beta$ in the x-axis and y-axis and showing rectangle area is smaller than area under the plots). Is there a proper algebraic ...
2
votes
1answer
130 views

How to show that $uv \le \frac {u^p}{p}+ \frac {v^q}{q}$ where $\frac 1p + \frac 1q = 1$ [duplicate]

Let $u, v, p, q > 0$, and let $\dfrac 1p + \dfrac 1q = 1$. Then $$uv \le \dfrac {u^p}{p}+ \dfrac {v^q}{q}$$ Show that equality occurs iff $u^p = v^q$. This is problem $10$ from Ch. 6 in ...
28
votes
7answers
2k views

Given $y_n=(1+\frac{1}{n})^{n+1}$ show that $\lbrace y_n \rbrace$ is a decreasing sequence

Given $$ y_n=\left(1+\frac{1}{n}\right)^{n+1}\hspace{-6mm},\qquad n \in \mathbb{N}, \quad n \geq 1. $$ Show that $\lbrace y_n \rbrace$ is a decreasing sequence. Anyone can help ? I consider the ...
10
votes
4answers
5k views

Binomial theorem proof for rational index without calculus

I have tried to find a proof of the binomial theorem for any power, but I am finding it difficult. One can obviously prove the integer index case using induction, but all of the approaches for ANY ...
5
votes
5answers
1k views

Prove that $xy \leq\frac{x^p}{p} + \frac{y^q}{q}$

OK guys I have this problem: For $x,y,p,q>0$ and $ \frac {1} {p} + \frac {1}{q}=1 $ prove that $ xy \leq\frac{x^p}{p} + \frac{y^q}{q}$ It says I should use Jensen's inequality, but I can't figure ...
7
votes
4answers
429 views

Find a limit in an efficent way

I'm trying to calculate the following limit: $$\mathop {\lim }\limits_{x \to {0^ + }} {\left( {\frac{{\sin x}}{x}} \right)^{\frac{1}{x}}}$$ What I did is writing it as: $${e^{\frac{1}{x}\ln \...
5
votes
4answers
223 views

Limit of $2^{n} n!/n^{n}$ as $n \to \infty$

Prove that the $\lim_{n \rightarrow \infty} \frac{2^{n} n!}{n^{n}} = 0$ $\rightarrow \frac{2^{n} n!}{n^{n}} = $ $(\frac{2}{n})^{n} n!$ Its possible to say that $\lim_{n \rightarrow \infty} $$\frac{...
2
votes
2answers
1k views

Proving that $\left(1 + \frac{x}{n}\right)^n, n \in \mathbb N$ is bounded

Please can you help me with the following question $$E(x) = \left\{ \left(1 + \frac{x}{n}\right)^n : n \in \mathbb N \right\}$$ Let a(x) = sup E(x) (least upper bound) without finding the sup of E(...
8
votes
3answers
171 views

How can I prove $\frac {d}{dx} {x^n} = n x^{n-1}$ for $ n \in \Bbb R$ without circular reasoning? [duplicate]

I just cannot prove that $$\frac {d}{dx} {x^n} = n x^{n-1}$$ for $ n \in \Bbb R$. For $n \in \Bbb{N}$, I can use the definition of a derivative : $$\frac {d}{dx}x^n = \lim_{h \rightarrow 0} \frac{(...
3
votes
2answers
927 views

Bounds of $f(x)=(1-ax)^{1/x}$

Let $a\in(0,1)$ be a fixed number. What is the numeric value of upper and lower bound of $f(x)=(1-ax)^{1/x}$ on $x\in (0,1)$? I feel as though I'm missing something, because it shouldn't be ...
2
votes
2answers
645 views

PROVE if $x \ge-1 $then $ (1+x)^n \ge 1+nx $ , Every $n \ge 1$

Use mathematical induction to prove this. Here is my answer but I stuck at certain point. Base Case: n=1 $$(1+x)^1 \ge 1+x $$ True , Induction Case: n=k assume $$(1+x)^k \...
2
votes
2answers
822 views

Proof that $\lim_{x\to \infty} (1+(x/n))^n = e^x$ Using Monotone Convergence Theorem

Use monotone convergence theorem to prove that $e(x):=\lim_{n\to\infty}(1+x/n)^n$ exists for all real $x$. Then show that $e(-x):=\lim_{n\to\infty}(1-x/n)^n=1/e(x)$. I'm struggling to use monotone ...
4
votes
4answers
144 views

Show that $ \left( 1 + \frac{x}{n}\right)^n$ is uniformly convergent on $S=[0,1]$. [duplicate]

Show that $ \left( 1 + \frac{x}{n}\right)^n$ is uniformly convergent on $S=[0,1]$. Given $f_n(x)=\left( 1 + \frac{x}{n}\right)^n$ is a sequence of bounded function on $[0,1]$ and $f:S \rightarrow \...
6
votes
2answers
317 views

Show sequence ${a_n} = \sqrt[n]{{{3^n} + {5^n}}}$ is monotone decreasing

(a) Show that sequence ${a_n} = \sqrt[n]{{{3^n} + {5^n}}}$ is monotone decreasing Proof Let ${a_n} = \sqrt[n]{{{3^n} + {5^n}}} = 5{\left[ {{{\left( {\frac{3}{5}} \right)}^n} + 1} \right]^{\frac{1}{n}...
2
votes
2answers
155 views

Binomial theorem and the inequality $(1-e^{-x})^{\alpha}\leq (1-\alpha e^{-x})$ for $0<\alpha\le 1$

Assume $0<\alpha\leq 1$ and $x>0$. Does the following inequality hold? $$(1-e^{-x})^{\alpha}\leq (1-\alpha e^{-x})$$ I know that the reverse inequality holds if $\alpha\ge 1$.

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