So I need to prove the inequality :  $$n &lt; 2^n$$  by Induction.  What I have done so far is :     Step $1$: Prove that the statement is true for $n=1$   $$1&lt;2^1$$ (true)      Step $2$:    Prove that, if $p(n)$ is true, then $p(n+1)$ is also true.      Assume that $p(n): n &lt; 2^n$ is true. then :      (Add $1$ on both side to get $n+1$) : $$n+1 &lt; 2^n +1$$      (Multiply the original inequality by $2$ to get $2^{n+1}$ ) :   $$2^{n+1} &lt; 2^n . 2$$   I don't know where to go from here. In fact, I am not sure whether what I did is true or not. I appreciate any insight and help.
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I am currently in the process of learning how to prove statements by induction. For the most part, I understand but I am at most clueless when it comes to prove some inequalities.  For example, I am having a hard time proving that: $n &lt; 2^n$ is true. (Even if it's clear that it is indeed true!)  In the induction step, I am stuck with: $n + 1 &lt; 2^{n + 1}$  I don't know what would be the assumption I should make from this statement.  Anyone could point me in the right direction? Thanks!
111670
I have got an inequality problem which is as follow:     Show that $e^n&gt;\frac{(n+1)^n}{n!}$   I can do it by induction but I have been told to prove it without induction.  My Work:  $$e^n=1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+........$$ $$e^n&gt;1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+........+\frac{n^n}{n!}$$ $$e^n&gt;\frac{n^n}{n!}+\frac{n^{n-1}}{(n-1)!}.......+\frac{n^2}{2!}+n+1$$  From here I can't go further.   I shall be thankful if you guys can provide me a complete solution/proof of this inequality. A hint will also work.  Thanks in advance.
8473
Problem  For what natural numbers is $n^3 &lt; 2^n$?  Attempt @ Solution   For $n=1$, $1 &lt; 2$ Suppose $n^3 &lt; 2^n$ for some $n = k \ge 1$ It looks like the inequality is true for $n = 0$, $n = 1$ and $n\ge10$ But, how can I prove this through induction?
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I was trying to prove this inequality using induction, but couldn't do.  Question: Suppose $a$ and $b$ are real numbers with $0 &lt; b &lt; a$. Prove that if $n$ is a positive integer, then:  $$a^n-b^n \leq na^{n-1}(a-b)$$
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I'm trying to do homework problems and for the most part I've been getting the results. For this one though, I am having some trouble since its $2^n$ and I can't relate it properly:     Prove using simple induction that $n &lt; 2^n$ for all $n \ge 1$.   So obviously, the basis step holds as $1 &lt; 3$. Now, I assume that $n = k$ holds as well and have to prove for $n=k+1$. This is the step I am having trouble with as I cannot relate the induction hypothesis with what I want to end up with. Can anybody show me a model solution for this one? I think my trouble comes because of the $2^n $
65335
How would I show that $P(\mathbb{N})$ is equivalent to $2^\mathbb{N}$? The questions asks to form a theorem from the statement I just gave. Any help would be appreciated.
101451
So I need to use induction to prove that $4^n&lt;5^n$ for all natural numbers $n &gt; 0$. I have proved the base case to be true. For the induction hypothesis, we assume that when $n=k$, then $4^k &lt; 5^k$. However, I am stuck in the inductive step when I try to show that it needs to be true for $n+1$. Therefore, we need to show that $4^{k+1} &lt; 5^{k+1}$. However I get stuck on the first step trying to prove it as in  $4^{k+1} = 4 \cdot 4^k$, then by substituting $5^k$ instead of $4^k$ from the inductive hypothesis and I get stuck right here and I do not know how to proceed.  Any help would be appreciated.
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.
88351
I would like to prove the following inequality, for $n=0,1,2,...$,   $$ \binom{2n}{n}\leq 4^n.$$ I already proved it by induction, and I'm looking for another proof.
115559
I have been trying to show these two inequalities hold for all positive integers n, but I don't know how to proceed at all... I have tried playing around with them but I didn't find anything helpful. I have also tried induction but I couldn't make it work.
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Use mathematical induction to show that:  $1^3 + 3^3 + 5^3 + ... + (2n + 1)^3 = (n+1)^2(2n^2+4n+1)$ whenever $n$ is a positive integer.  I am able to solve this problem, but the thing that confuses me on this is the Basis Step.  P(1): $(2(1) + 1)^3$ $\neq$ $4(7)$, does this mean that this problem can not be proved using mathematical induction, or am I missing something? Thanks.
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I am currently helping a friend of mine with his preperations for his next exam. A big topic of the exam will be induction, thus I told him he should practice this a lot. As at the beginning he had no idea how induction worked, I showed him some typical examples.  Now he showed me an exercise he was having trouble with, which states that one should prove $$n^{n+1} &gt; (n+1)^n$$  for all $n \geq 3$. I have to admit that I too have trouble showing this inequality, as in all of my attempts, my lower bound is too low. Also, I have not yet figured out, how one gets the $(n+2)^{n+1}$, primarely the number $2$ is a problem. I think this might be solved using the binomial theorem, however, I don't think they have already seen the binomial theorem in school.  Is there an easy method to show this inequality by induction not using the binomial theorem? If no: How can one show it using the binomial theorem?  Thanks for answers in advance.
31715
I'm trying to prove by induction the following statement without success: $$\forall n \ge 2, \;\forall d \ge 2 : d \mid n(n+1)(n+2)...(n+d-1) $$  For the base case: $n = 2$, $d = 2$ $2\mid 2(2+1)$ which is true.  Now, the confusion begins! I assume I would need to use the second induction principle to proof this because $P(n)$ and $P(n+1)$ are not related at all. It is also the first time I am dealing with more than one variable so it makes it harder for me.  I tried the following: - Trying to prove by simple induction. I did not go very far. - Trying to split my induction step in 2 parts: If $d\mid n$, I'm done.   If $d$ does not divide $n$, then I would need to do a second proof and this is where I'm blocked.  Anyone could tell me what's wrong in the approach I take to solve this problem?  Any help would be appreciated!
85800
I'm having trouble proving this by induction. We need to show that $P(k+1)$ is true: $$\sum_{i=2}^{k+1} \frac{1}{i^2}\leq 1.$$  Don't know where to go from here. Any help?
90221
Show that $2^n&gt;n$ for $n\in \mathbb{N}$   I can solve this by mathematical induction. Is there any other method to solve.
3368
Prove by mathematical  induction that:   $$\forall n \in \mathbb{N}: 3^{n} &gt; n^{3}$$  Step 1: Show that the statement is true for $n = 1$:  $$3^{1} &gt; 1^{3} \Rightarrow 3 &gt; 1$$  Step 2: Show that if the statement is true for $n = p$, it is true for $n = p + 1$  The general idea I had was to start with $(p+1)^{3}$ and during the process substitute in $3^{p}$ for $p^{3}$ as an inequality.  $$(p+1)^{3} = p^{3} + 3p^{2} + 3p + 1 &lt; 3^{p} + 3p^{2} + 3p + 1$$  Now, if it can be shown that:  $$\forall p \in \mathbb{N}: 3p^{2} + 3p + 1 \leq 2 \cdot 3^{p}$$  ...the proof is complete. This is because $3^{p+1} = 3 \cdot 3^{p}$ and one of those three have already been used.   We do this by mathematical induction. First, the base case of $n = 1$:  $$3\cdot 1^{2} + 3 \cdot 1 + 1 \not \leq 2 \cdot 3^{1}$$  ..which turns out to be false.  What are some more productive approaches to this step?
110000
How can I prove that $n!&lt;n^n$ for every $n&gt;1$ using the induction principle?  If I put $n=2$ I get $2&lt;4$ so I know that $n!&lt;n^n$ is true.  Now I don't know how to prove that $(n+1)!&lt;(n+1)^{n+1}$. I can write $(n+1)!=n!(n+1)$ but then I don't know how to go on.