Prove that ${\sqrt {n} }^{\sqrt {n+1}} &gt; {\sqrt {n+1}}^{\sqrt {n}};n=7,8,9...$ SOURCE : &quot;Inequalities proposed in CRUX Mathematicorum&quot;  I tried induction, but could not prove it. Can anybody provide a hint ?
123674
Which is greater? $\sqrt{n}^{\sqrt{n+1}}$ or $\sqrt{n+1}^\sqrt{n}$  I know that $\sqrt{n}^{\sqrt{n+1}}$ is greater but I tried using induction and I couldn't figure it out. Thanks for the help.
19128
My question is regarding the question posed at the end of the proof.  My answer is that the result does not hold for all $m \ge 7$ because when $m=7$, the result is $343 \le 128$, which is false.  Is there another way to answer this sort of question besides directly trying different values until you get a false result?  EDIT:  Another question I have regarding this solution is:  In the proof for the inductive step, we start by assuming $k \ge 10$. But along the way, the author mentions $k \ge 1$ and $k \ge 7$ to justify the inequality.  Why do we bother to do this instead of just sticking with $k \ge 10$?
54219
I'm not too sure exactly how to approach this question. Would anyone be able to give me any helpful advice or some sort of direction? I have a little problem with induction.   Prove by induction that:  $$1 + 3 + 5 + 7 +...+ (2n+1) = (n+1)^2 \quad\forall n \in \mathbb N $$
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Proove by the Complete Induction for every $n\in \mathbb{N}, n \geq 1$ $$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{n}} &gt; \sqrt{n}$$.  I know only the basis - $\frac{1}{\sqrt{1}} \geq \sqrt{1}$.  But, I dont know how to do the change from the regular induction to the induction of Set theroy(Complete Induction).
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I've done the base step, but how do I prove it is true for $n+1$ without using a fallacy? $$4(n!)&gt;2^{n+2}\quad \text{for } n\geq 4$$ Please help.
92281
Let $x_1 = 2$, and define $$x_{n+1} = \frac{1}{2} \left(x_n + \frac{2}{x_n}\right).$$ Show that $x_n^2$ is always greater than or equal to $2$, and then use this to prove that $x_n − x_{n+1} ≥ 0$. Conclude that $\lim x_n = \sqrt2$.   My question:  So I know how to do this problem but I don't know how to prove that $x_n &gt;0$ for all $n\in \mathbb{N}$.
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I'm having some trouble figuring out where to start for this proof.     Prove for all positive integers $n$,   $$\sqrt{n+1}-\sqrt{n}\lt \frac{1}{2\sqrt{n}}$$   Any help will be greatly appreciated.
109068
Find all positive integers $n$ such that $n$ is divisible by all the positive integers less than or equal to $\sqrt{n}$  My thought: If n is a positive integer, let d(n) denote the number of positive divisors of n and I will find n such that $d(n) \ge 2\sqrt{n} - 1$
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Problem Prove that $n! &gt; \sqrt{n^n}, n \geq 3$.    I'm currently have two ideas in mind, one is to use induction on $n$, two is to find $\displaystyle\lim_{n\to\infty}\dfrac{n!}{\sqrt{n^n}}$. However, both methods don't seem to get close to the answer. I wonder is there another method to prove this problem that I'm not aware of? Any suggestion would be greatly appreciated.
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i have a pretty nasty question. i was glancing through a few olympiad papers and stumbled upon this question:  prove that  $ i = \sqrt {-1}\ $.  i tried the conventional methods namely euler's formula  but could not figure what to do next. how do you actually proceed further? i have checked a few books and a proof is not available.are there any proof available for this
23125
I aim to show that $$(n+1)^n&lt;n^{(n+1)}$$ for all $n \geq 3$. I tried induction, but it didn't work. What should I do?
123132
I am having difficulty proving the following inequality: $$ \sqrt[n]{n} &lt; 1 + \sqrt{\frac{2}{n}} \quad \text{for all positive integers}\,\,\, n. $$ I am trying to use mathematical induction but I am having trouble going from the left side to the right side (in the induction step) and vice versa.   I also tried to take the power of $n$ on both sides and somehow use the binomial expansion but I am having difficulty with that. Thank you!
123925
I'm struggling around my homework. I hope someone will point me the right direction for solving following examples:   Prove that $n^{n+1} &gt; (n + 1)^n$ for $n &gt; 2$; Prove that $(1 + x)^n \ge 1 + nx$; $x \in\Bbb R$; $n \in\Bbb N$; Prove that $(2n)! &lt; 2^{2n}(n!)^2$; $n \ge 1$; Prove that $2^n &gt; n$:   Thank you a lot.
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Prove that  $ n &lt; 2^{n}  $ for all natural numbers $n$.  I tried this with induction: Inequality clearly holds when $n=1$. Supposing that when $n=k$, $k&lt;2^{k}$. Considering $k+1 &lt;2^{k}+1$, but where do I go from here?  Any other methods maybe?
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$n$ is a natural number. I’ve tried to turn this problem to this: Prove that: $n(n+1)(n+2)(n+3)+1 $is always a perfect square. So: $$n(n+1)(n+2)(n+3)+1 =x^{2}$$ And i’ve expanded the whole equation but it didn’t help me a lot.
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Assumption: $\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}\ge\frac{2}{3}k\sqrt{k}$   Prove true for $n=k+1$ $$\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}+\sqrt{k+1}\ge\frac{2}{3}(k+1)\sqrt{k+1}$$  I'm upto :$$\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}+\sqrt{k+1}\ge\frac{2}{3}k\sqrt{k}+\sqrt{k+1}$$after which I'm stuck.
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This question is from Apostol's Mathematical Analysis. I have a solution for it at   The question is:  Prove that $\sqrt {n − 1} +\sqrt {n + 1}$ is irrational for every integer n ≥ 1.  And the proof I have is  Now, I can't understand how to get h=$3/2$ and k=$1/2$ ?  Please help.