Here is an inequality (let's call it - "$A(n)$") that has to be proved: $$  (n+1)^n &lt; n^{n+1} \text{ for } n\ge3. $$  I'll skip the first two steps of induction and move right to the induction step (step #3) in order not to waste time.   Assume that $n = k+1$. I get: $$ (k+2)^{k+1} &lt; (k+1)^{k+2} $$ From here: $$ (k+2)^k(k+2) &lt; (k+1)^k(k+1)^2 $$ and that is where i'm stuck.   I've found, that the inequality $A(n)$ (by dividing it by $n^n$) can be expressed as: $$ (1+1/n)^n &lt; n. $$ Here again I substitute $n = k+1$ and get: $$ (1+1/(k+1))^{k+1} &lt; k+1. $$ Stuck here as well...  On the forum I've found ways to prove the inequality by Binomial theorem and other approaches, but how to prove this with math induction ? It would be great if answers included comments on every step, that is done to find the solution...
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.
70085
Problem: $2^n &gt; n^2, \forall n \in \mathbb{N} , n \geq 5$ Base: $2^5 &gt; 5^2$ Induction Hypothesis: Assume for $n = k \geq 5$ that $2^k&gt;k^2$ Inductive Step: $$2^k &gt; k^2$$ $$2^k \times  2 &gt; k^2 \times 2$$ $$2^{k+1} &gt; 2k^2$$ From there I can finish the proof by asserting that $k^2 &gt; 2k+1, \forall k \in \mathbb{N} , k \geq 3$. Do I need to prove that $k^2 &gt; 2k+1, \forall k \in \mathbb{N} , k \geq 3$. Or can I substitute it into my inequality?
29081
I want to prove by induction that $a^n \mid b^n$ implies that $a \mid b$ holds for all integers $n\geq 1$.  Clearly for $n=1$ this is true, since if $a \mid b$, then $a \mid b$.  Suppose this is true for some $n = k$. Then $a^k \mid b^k$, so $a|b$.  $a^k \mid b^k$ means there exists some integer $m$ such that $b^k = ma^k$, and $a \mid b$ means there exists an integer $r$ such that $b=ra$.  Then we can multiply both sides of the above equality by any constant, namely $b$.   Then $b \cdot b^k = bma^k$  so $b^{k+1} = (ra)ma^k = rma^{k+1}$  which means that $a^{k+1} \mid b^{k+1}$ completing the induction step, and thus proving that for all integers $n \geq1$ the original statement holds.  Are there any logical holes in the proof?
10772
I'm told to show that $2^n &lt; n!$ using induction  This is my attempt at it:  BC: $n=4, 2^4 = 16 &lt; 4!$  IH: n = k, $2^k &lt; k!$  IS: try n = k+1  I'm told to only work from one side, so I try the left side:  2^(k+1)   But I'm stuck here, any ideas?
121674
Prove that $2n-3\leq 2^{n-2}$ , for all $n \geq 5$ by mathematical induction  I have to prove by mathematical induction that: $2n-3\leq 2^{n-2}$ , for all $n \geq 5$  Thank you for the Review.
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:  $$1 + 3 + 5 + 7 +...+ (2n+1) = (n+1)^2 \quad\forall n \in \mathbb N $$
116474
I'm new to inequalities in mathematical induction and don't know how to proceed further. So far I was able to do this:  $V(1): 1≤1 \text{  true}$  $V(n): n!≤((n+1)/2)^n$  $V(n+1): (n+1)!≤((n+2)/2)^{(n+1)}$  and I've got : $(((n+1)/2)^n)\cdot(n+1)≤((n+2)/2)^{(n+1)}$ $((n+1)^n)n(n+1)≤((n+2)^n)((n/2)+1)$
82812
Prove that $$3^n &gt; n^2$$  I am using induction and I understand that when $n=1$ it is true. The induction hypothesis is when $n=k$ so $3^k&gt;k^2$. So for the induction step we have $n=k+1$ so  $3^{k+1} &gt; (k+1)^2$ which is equal to $3\cdot3^k &gt; k^2+2k+1$. I know you multiple both sides of the induction hypothesis by $3$ but I'm not sure what to do next.
29302
How would you go about proving $n! &lt; n^n$ using a mathematical induction proof?  I understand how to solve inductive proofs with = but I'm getting a bit lost in this example.   Any help is much appreciated.
14809
Let $n$ and $1\leq k \leq n$ be natural numbers. Prove the inequality     $$\sum_{i=k}^n \binom{n}{i}\bigg(\frac{k}{n+1}\bigg)^i\bigg(1-\frac{k}{n+1}\bigg)^{n-i} \leq 1 - \frac{1}{e} $$  Equivalently, if $X\sim$ Bin($n$,$\frac{k}{n+1}$), prove that $\mathbb{P}[X\geq k] \leq 1 - \frac{1}{e}$.   My attempt: It may be helpful to show that the LHS tends to $1-\frac{1}{e}$ as $n \to \infty$ (already did that) and that the LHS is an increasing function on $n$ (have not done that).
73031
Let $(a_n)_{n \in \mathbb{N}}$ a sequence of real numbers with $0 \leq a_n \leq 1$ for all $n \in \mathbb{N}$.  I want to prove the following inequality using mathematical induction:  $\prod \limits_{k=0}^{n}(1-a_k) \geq  1- \sum\limits_{k=0}^{n}a_k$.  It's obviously true for $n=0$.  $n \rightarrow n+1$  Assume that it's true for n. Then one has:  $\prod \limits_{k=0}^{n+1}(1-a_k) = (1- a_{n+1}) \prod \limits_{k=0}^{n}(1-a_k) \geq (1- a_{n+1}) (1- \sum\limits_{k=0}^{n}a_k) = 1- \sum\limits_{k=0}^{n}a_k - a_{n+1} + a_{n+1} \sum\limits_{k=0}^{n}a_k = 1- \sum\limits_{k=0}^{n+1}a_k + a_{n+1} \sum\limits_{k=0}^{n}a_k=...$  How to go on?
18135
How do I prove that $3^n &gt; n^2$ with mathematical induction? I thought I had the correct answer but my teacher says its wrong.  I let $n=1$ for the initial case and it works. I then assumed $n=k$ works and went onto the $n=k+1$ case, but after that it goes a little awry. Thanks in advance!
15822
I am trying to prove: $P(n): |x_1| + \cdots + |x_n| \leq |x_1 + \cdots +x_n|$ for all natural numbers $n$. The $x_i$ are real numbers.  Base: Let $n =1$:    we have $|x_1| \leq |x_1|$ which is clearly true  Step: Let $k$ exist in the integers such that $k \geq 1$ and assume $P(k)$ is true.  This is where I am lost. I do not see how to leverage the induction hypothesis.  Here is my latest approach:  Can you do the following in the induction step: Let $Y$ = |$x_1$ +...+$x_n$| and Let $Z$ = |$x_1$| +...+ |$x_n$| Then we have: |$Y$ + $x_n+1$| $\leq$ $Z$ + |$x_n+1$|. $Y$ $\leq$ $Z$ from the induction step, and then from the base case this is just another triangle inequality. End of proof.
122397
Donald Knuth &quot;The Art of Computer Programming&quot; Section 1.2.1 has an exercise No.2, the solution to which I don't understand. Exercise:  There must be something wrong with this proof. What is it? Theorem. Let $a$ be any positive number. For all positive integers $n$ we have $a^{n-1}=1$. Proof. If  $n = 1, a^{n-1}=a^{1-1}=a^0=1$. And by induction, assuming that the theorem is true for $1, 2, ..., n$, we have $a^{(n+1)-1}=a^n=\frac{a^{n-1}\times a^{n-1}}{a^{(n-1)-1}}=\frac{1\times1}{1}=1$; so the theorem is true for $n+1$ as well.  As far as I understand the proof by induction, the solution should be approximately following:$a^{1-1}=1$Assume that $a^{n-1}=1$;Therefore $a^{(n+1)-1}=1\times a$;So $a^{(n+1)-1}$ does not equal $a^{n-1}$ and does not equal $1$, unless $a=1$ The solution to exercise reads:  The theorem has not been proved for $n=2$; in the second part of the proof, take $n=1$; we assume there that $a^{-1}=1$. If this condition is true (so that $a=1$), the theorem is indeed valid.  I want to compare it to another, correct, proof by induction in the example from :  Adding up Odd Numbers $1+3+5+...+(2n−1)=n^2$  Show it is true for $n=1$ $1=1^2$ is True Assume it is true for $n=k$ $1+3+5+...+(2k−1) = k^2$ is True (An assumption!) Now, prove it is true for $(k+1)$ $1 + 3 + 5 + ... + (2k−1) + (2(k+1)−1) = (k+1)^2$We know that $1 + 3 + 5 + ... + (2k−1) = k^2$ (the assumption above),so we can do a replacement for all but the last term:$k^2 + (2(k+1)−1) = (k+1)^2$Now expand all terms:$k^2 + 2k + 2 − 1 = k^2 + 2k+1$And simplify:$k^2 + 2k + 1 = k^2 + 2k + 1$They are the same! So it is true.So: $1 + 3 + 5 + ... + (2(k+1)−1) = (k+1)^2$ is True   Returning to the first example, I didn't quite understand the meaning of the phrase  The theorem has not been proved for $n=2$;  In the second (correct) proof there is no proving that it holds for $n=2$, despite covering the same topic. And Knuth himself has the proof of the theorem $1+3+5+...+(2k−1) = k^2$ similar to the one above, where he too doesn't prove it for $k=2$. So what did he mean by this phrase? Does he mean that there already is an exception in the rule stated when we take arbitrary $n$, for example $n=2$, so therefore it wouldn't hold for all examples anyway? But taking arbitrary numbers and putting it into equation instead of $n$ doesn't seems to be an effective and systematic way to debunk such proofs. How would you explain what is wrong with the exercise?
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.
24893
Prove by induction that $\forall n \in\mathbb N$, $3\mid4^{n}-1$.  1) For $n = 1$ the statement is obviously true.  2) Now what about $n+1$? I was thinking of writing $4^{n}-1$ as $2^{2n}-1$ and then $4^{n+1}-1 = 2^{2n+2}-1$ but that got me nowhere.
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.