Let $(2n)!!$ be the product of all positive even integers less than or equal to $2n$. Let $(2n − 1)!!$ be the product of all odd positive integers less than or equal to $(2n − 1)$. Prove that  $$\frac{(2n − 1)!!}{(2n)!!} \leq \frac{1}{\sqrt{2n+1}}.$$ I am up to the inductive step, where am I stuck. I can't find a way to substitute in from the assumption because of the square root.
53994
Problem statement: Prove that $\frac{1*3*5*...*(2n-1)}{2*4*6*...(2n)}\leq \frac{1}{\sqrt{2n+1}}$ and that there exists a limit when $n \to \infty $. , $n\in \mathbb{N}$   My progress  LHS is equivalent to $\frac{(2n-1)!}{(2n)!}=\frac{(2n-1)(2n-2)(2n-3)\cdot ....}{(2n)(2n-1)(2n-2)\cdot ....}=\frac{1}{2n}$ So we can rewrite our inequality as:  $\frac{1}{2n}\leq \frac{1}{\sqrt{2n+1}}$ Let's use induction:  For $n=1$ it is obviously true. Assume $n=k$ is correct and show that $n=k+1$ holds.  $\frac{1}{2k+2}\leq \frac{1}{\sqrt{2k+3}}\Leftrightarrow 2k+2\geq\sqrt{2k+3}\Leftrightarrow 4(k+\frac{3}{4})^2-\frac{5}{4}$ after squaring and completing the square. And this does not hold for all $n$  About convergence: Is it not enough to check that $\lim_{n \to \infty}\frac{1}{2n}=\infty$ and conclude that it does not converge?
111685
I need to prove  $2n \leq 2^n$, for all integer $n≥1$ by mathematical induction?  This is how I prove this:  Prove:$2n ≤ 2^n$, for all integer $n≥1$  Proof: $2+4+6+...+2n=2^n$  $i.)$ Let $P(n)=1         P(1): 2(1)=2^1\implies 2=2$.        Hence, $P(1)$ is true.  $ii.)$  Assume that $P(n)$ is true for $n=k$, i.e, $2+4+6+...+2k=2^k$, and prove that $P(n)$ is also true for $n=k+1$, i.e, $2+4+6...+2(k+1)=2^{(k+1)}$  from the assumption add $2(k+1)$ on both sides so we have   $2+4+6...2k+2(k+1)=2^k+2(k+1)$  I'm confused with $2^k+2(k+1)$, I don't know how to make $2^k$ be equivalent to $2^{k+1}$.  I feel i'm doing something wrong.   Any help would be appreciated!
113910
EDIT:  My question was poorly worded.   I wasn't asking about showing $\ln(1+x) &gt; \frac{2x}{2+x}$ for $x&gt;0$.   What I wanted to know is why the lower bound provided by $ \frac{2x}{2+x}$ was so tight for small positive values of $x$. This is addressed in robjohn's answer.
118824
Prove that for any positive integer $n$, the number of odd integers among the binomial coefficients $\binom{n}{k}$ $(0 \leq k \leq n)$ is a power of $2$.   I am thinking of trying to prove that the number of odd binomial coefficients is equal to 2 to the power of the number of $1$s in the binary form of $n$. How should I go about proving this?
110079
Prove that $$x^n-1=(x^2-1)\prod_{k=1}^{(n-2)/2}[x^2-2x\cos{\frac{2k\pi}{n}}+1]$$ if $n$ be an even positive integer. Hence deduce that $$\sin{\frac{\pi}{32}}\sin{\frac{2\pi}{32}}\sin{\frac{3\pi}{32}}\cdots\sin{\frac{15\pi}{32}}=\frac{1}{2^{13}}$$  Attempt:  I am able to solve the 1st part of the problem and unable to solve the 2nd part that is the deduction part of the problem. Please help for the deduction part only.   Edit:  Using double angle: I can show that $$\sqrt{n}=2^{(n-1)/2}\sin{\frac{\pi}{n}}\sin{\frac{2\pi}{n}}\cdots \sin{\frac{(n-2)\pi}{2n}}$$  Please suggest suitable n and the please help me to complete the remaining part. I have taken n=32 but not able to complete the solution.
36572
I would like to show that $\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\leq \frac{1}{\sqrt{3n+1}}$ holds for all natural numbers. I got stuck here:  $\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\cdot\frac{2n+1}{2n+2}\leq \frac{1}{\sqrt{3n+1}}\frac{2n+1}{2n+2}.$  I would appreciate your help.
61352
If $n$ be a positive integer greater than $1$, then prove that $$2^n&gt;1+n\sqrt{2^{n-1}}$$    I found this problem under $AM \ge GM$ chapter. Help me to solve this problem using  $AM \ge GM$. Thanks in advance.
109289
For positive integers $m,n$ if $\sqrt 7  - \frac{m}{n} &gt; 0$ then prove that $\sqrt 7  - \frac{m}{n} &gt; \frac{1}{{mn}}$.
106305
It is straight forward to show that $\sqrt[n]{n!} &lt; n$ from:   $2 &gt; \sqrt{2}$ $\frac{n+1}{n} &gt; \frac{\sqrt[n+1]{n+1}}{\sqrt[n]{n}}$ since $\frac{n+1}{\sqrt[n+1]{n+1}} &gt; \frac{n}{\sqrt[n]{n}}$    Is this the best that can be done?  Is it possible to find a better upper bound for $\sqrt[n]{n!}$
85098
I need help!  I need to prove that for any $2 \le n,k$ positive integers $(n - 1)^2 \mid n^k -1$ if and only if $(n - 1) \mid k$  Thanks!
64569
I have to show that if $|z| &lt; 1$, $z \in \mathbb{C}$, $$(1+z)\prod_{n=1}^{+\infty}(1+z^{2^{n}})= \frac{1}{1-z}$$  I want to understand how to do this kind of exercises, any hint ?
67668
I've been struggling with this all day today. I imagine it's not very hard, but my algebra skills are terrible. So, how can I show that if $m&gt;n$ and $a$ is a positive integer, then $$a^{2^n}+1 \mid a^{2^m}-1.$$ I just can't get a coherent picture of what I'm supposed to do. If this too (see yesterday's question) boils down to the Unique Factorization Theorem, I would be very grateful for any intuition on how to think about this!
104310
I need to prove that for every natural n that happens and the prove needs to be by induction, I wont prove for n=1 because its obvious,so lets move to n and n+1 $$ \frac{(2n!)}{(n!)^2} ≥ \frac{(4^n)}{(2n+1)} $$ the assumption for n+1 $$ \frac{(2n+2)!}{(n+1)!^2} ≥ \frac{(4^{n+1})}{2n+3} $$ now I dont now how to progress,I tried this for n+1 $$ \frac{(2n+2)(2n+1)(2n!)}{(n+1)!^2} ≥ \frac{(4^{n}4^1)}{(2n+1)+2)} $$ and had a suggestion to use $ 2n! = (n!) \prod _{n=n}^{2n}\:n\cdot )! $ but cant see how it helps,ty.
39946
Find all $n&gt;1$ such that $\dfrac{2^n+1}{n^2}$ is an integer.   I know that $n$ must be odd, then I don't know how to carry on. Please help. Thank you.
74594
Using permutation or otherwise, prove that $\displaystyle \frac{(n^2)!}{(n!)^n}$ is an integer,where $n$ is a positive integer.  I have no idea how to prove this..!!I am not able to even start this Can u give some hints or the solution.!cheers.!!
60825
Let $a$, $b$, $c$, $d$ and $e$ be positive numbers. Prove that:  $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+e}+\frac{e}{e+a}\geq\frac{a+b+c+d+e}{a+b+c+d+e-3\sqrt[5]{abcde}}$$  Easy to show that $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$$ is true  and for even $n$ and positives $a_i$ the following inequality is true.  $$\frac{a_1}{a_1+a_2}+\frac{a_2}{a_2+a_3}+...+\frac{a_n}{a_n+a_1}\geq\frac{a_1+a_2+...+a_n}{a_1+a_2+...+a_n-(n-2)\sqrt[n]{a_1a_2...a_n}}$$
24256
Given that $x_n = \displaystyle \prod_{i=1}^n  \frac{2i-1}{2i}$  Then prove that $x_n \leq \frac{1}{\sqrt{3n+1}}$ for all $n \in \mathbb Z_+$  What I did was take the logarithm of $x_n$, and I arrived at: $\log{x_n}=\displaystyle \sum_{i=1}^{n} (\log{(2i-1)} - \log{2i}) $  I'd like to know if I proceeded correctly, and thus would like further guidance to solve the problem. However, if I haven't approached the problem correctly, I'd appreciate hints and techniques that are applicable. Please don't post the whole answer because I'd like to work this out on my own. Thanks.