1
vote
4answers
41 views

Prove $\forall n \in\Bbb N$, $0 < a < 1$ $\implies$ $a^n \leq 1$

I'm trying to prove this by induction but I'm running into some trouble. The base case is $0$, so, $a^0 = 1$, the inequality holds true Being new to induction, I don't exactly know what to do for ...
1
vote
2answers
70 views

Proving inequality $3^{n^2} > (n!)^4$

Prove that $3^{n^2} > (n!)^4$ for all positive integers $n$. I tried to use induction on this problem but failed to do so. I instead tried to prove $3^{2n+1}>(n+1)^4$, but couldn't come up ...
4
votes
1answer
74 views

how to solve this elementary induction proof: $\frac{1}{1^2}+ \cdots+\frac{1}{n^2}\le\ 2-\frac{1}{n}$

this is a seemingly simple induction question that has me confused about perhaps my understanding of how to apply induction the question; $$\frac{1}{1^2}+ \cdots+\frac{1}{n^2}\ \le\ 2-\frac{1}{n},\ ...
0
votes
1answer
39 views

Inductive proof of inequality $a\le ab$ for nonnegative integers

I reading about of proof of the claim "If $a \ge 0$ and $b > 0$, then $a \le ab$. (Here $a$ and $b$ are integers.) The proof the author is employing is inductive. I understand the basis case; ...
3
votes
3answers
41 views

Prove that $2n+1 \leq 2^n$ for $n \geq 3$ using mathematical induction.

Question: $2n+1 \leq 2^n$, for all $n \geq 3$ I've tried: Basis: $P(3) = 7 \leq 8 $, so basis step is valid Pick an arbitrary value from the universe, $k \geq 3$ Inductive Step: $2k + 1 \leq ...
1
vote
2answers
54 views

How can I show that $n^{n+2}<(2n)!$ for any integer $n$.

When I was try to show that the series $\sum_n \frac{n^n}{(2n)!}$ is convergent using comparison test, I stuck at the point $n^{n+2}<(2n)!$ I think it can be show using mathematical induction. If ...
2
votes
1answer
33 views

is my induction proof sufficient?

question; prove that $\forall\ n\ge4, n\in \mathbb{Z}, \ n!\gt n^2$. my work; let $n=4$ then $4!=24 \gt 4^2=16.$ true. now assume $n! \gt n^2$ is true for all $n\le k$ so now assume $k! \gt ...
0
votes
3answers
54 views

In proof by induction, what does it mean when condition for inductive step is lesser than the propsition itself?

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. ...
1
vote
1answer
54 views

How to use mathematical induction with inequality?

I am stuck with this question. Given that $n$ is a positive integer where $n≥2$, prove by the method of mathematical induction that (a) $$ \sum_{r=1}^{n-1} r^3 < \frac{n^4}{4} $$ (b) $$ ...
2
votes
1answer
44 views

What is the most elementary proof of these inequalities?

Let $p$ be a non-zero integer, and let $x_1$, $\ldots$, $x_n$ be $n$ positive real numbers. Then we define the $p$-th power mean $M_p$ of these numbers as $$ M_p \colon= (\frac{x_1^p + \ldots + ...
0
votes
3answers
52 views

How to derive this inequality?

How to derive the following inequality for all positive integers $n \geq 2$? $$ \frac{n!}{n^n} \leq \left(\frac{1}{2}\right)^k,$$ where $k$ denotes the greatest integer less than or equal to $\dfrac ...
0
votes
2answers
40 views

How to derive these inequalities?

I can derive the inequalities $$ n^p < \frac{(n+1)^{p+1} - n^{p+1}}{p+1} < (n+1)^p $$ for any positive integers $p$ and $n$. These actually follow from the identity $$b^p - a^p = (b-a)(b^{p-1} + ...
0
votes
2answers
63 views

I can prove that the series is greater than $\frac{1}{2}$ however i can't prove that it is greater than $\frac{13}{24}$ [duplicate]

Prove that for any positive integer $n>1$ $$ \frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} \ldots + \frac{1}{2n} > \frac{13}{24} $$ I can prove that the series is greater than $\frac{12}{24}$ ...
1
vote
1answer
64 views

Prove $x_n \leq x_{n+1}$ for all $n$ by induction

Prove $x_n \leq x_{n+1}$ for all $n$ by induction. I am reading this example from "Understanding Analysis" by Abbott (page 10). He says the multiple across the inequality by $1/2$ and then add 1 to ...
0
votes
4answers
49 views

$(n+1)!>n^2, \forall n\ge 4$

The base case is clear, since if $n\gt 4, 5!=120\gt 25=5^2$ So assume $n=k$ which shows that $k!\gt k^2$. Then if $n=k+1$, $$(k+1)!=(k+1)k!$$ $$\gt (k+1)k^2 $$ Induction argument $$=k^3+k^2$$ $$\gt ...
2
votes
5answers
104 views

Hint in Proving that $n^2\le n!$

The problem is to find the values of n such $n^2\le n!$. I have found this set to be {${n\in Z^+| n\le1 \lor n\ge 4}$} I've used a proof by cases to prove the first part ($\forall n\le 1$) which was ...
2
votes
3answers
56 views

Proof by induction of a recursive sequence

I am studying CIE A levels Further Maths and I am stuck at a question from June 2002: Q The sequence of positive numbers $u_1,u_2,u_3,...$ is such that $u_1<4$ and ...
1
vote
1answer
50 views

Strong induction inequality proof

Use strong induction to prove that $$\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{n^3}\leq\frac{5}{8}-\frac{1}{n}$$ $$n\geq2$$ I'm not sure how to go about this. I used base cases n=2, and n=3 but ...
0
votes
1answer
33 views

How to use induction on this type of inequality?

Given $a_1,a_2,\ldots,a_n>0$ and $a_1+a_2+\ldots+a_n<\frac{1}{2}$, prove that $(1+a_1)(1+a_2)\ldots(1+a_n)<2$. Some of you may have already seen this inequality. I was the one who asked ...
0
votes
1answer
48 views

Induction proof that $4^n > 3^n+2^n$ for $n\ge2$

This is a problem with induction and proofs but I'm not sure how to start with proving this one. $$\text{Show that for any $n \geq 2$, $4^n > 3^n+2^n$}$$
8
votes
2answers
47 views

How to prove an inequality

$a$, $b$, $c$, $d$ are rational numbers and all $> 0$. $\max \left\{\dfrac{a}{b} , \dfrac{c}{d}\right\} \geq \dfrac{a+c}{b+d}\geq \min \left\{\dfrac{a}{b} , \dfrac{c}{d}\right\}$ Hope someone ...
1
vote
4answers
188 views

Induction proof of $n^{(n+1) }> n(n+1)^{(n-1)}$

The question statement from my homework booklet goes: Prove by mathematical induction that $n^{n+1} > n(n+1)^{n-1}$ is true for all integers $n \geq 2$. I've managed to come up with this ...
2
votes
1answer
162 views

Proving that $\,\sqrt [n] n < 1 + \sqrt{\frac{2}{n}}\,$ for all positive $n$

Hello I am having difficulty proving the following inequality: $$ \sqrt[n]{n} < 1 + \sqrt{\frac{2}{n}} \quad \text{for all positive integers}\,\,\, n. $$ I am trying to use mathematical induction ...
0
votes
2answers
58 views

Using induction to prove $2^{n-1}(1 + a_1a_2\ldots a_n) \geq (1+a_1)(1+a_2)\ldots(1+a_n)$ for $a_i \geq 1$

Hello I have been blasting at this inequality proof and it is just not doing what I want it to do: Prove that $2^{n-1}(a_1a_2\ldots a_n + 1) \geq (1+a_1)(1+a_2)\ldots(1+a_n)$ assuming that ...
2
votes
1answer
35 views

Inductive proof of an inequality

I am trying to prove this inequality by induction: For all $x$ in the interval $x\in [0, \pi]$, prove that: $$ |\sin (nx)| \leq n\sin(x) \textit{, n a nonnegative integer}$$ The base case is ...
3
votes
3answers
136 views

Prove that, for any positive integer n: $(a + b)^{n} \leq 2^{n-1}(a^{n}+b^{n})$

Prove that, for any positive integer n: $(a + b)^{n} \leq 2^{n-1}(a^{n}+b^{n}) $ I tried induction theorem, when $n = 1$ it is obviously right. But, say $n=k$, It does not make sense since I cannot ...
1
vote
5answers
372 views

Invalid induction proof?

Prove the following using mathematical induction. If $a_{1}, a_{1}, ... , a_{n}$ are positive real numbers such that if $$a_{1}a_{2}...a_{n} = 1 $$ then $$a_{1}+a_{2}+ ... + a_{n} \geq n$$ My proof: ...
1
vote
1answer
68 views

Stuck on inductive step: $2^x > x^n$ when $x\rightarrow \infty$

I want to show that $2^x > x^n$ when $x \rightarrow \infty$ for all $n \in \mathbb{N}$. I'm trying to do it by induction over $n$. The base case, $n = 1$, is true: $2^x > x$ when $x \rightarrow ...
0
votes
1answer
71 views

Proving an inequality for a sequence by induction

I'm having some trouble with the following problem: Let $a_n$ be a sequence defined iteratively for $n \geq 0$ as follows: $a_n = a_{m+1} + 2a_m + a_{n-m-1} + 2$ where $m$ is defined as ...
3
votes
1answer
166 views

Proof by induction - correct inductive step?

The problem: $$ x_1 \geq x_2 \geq ... \geq x_{3n} \geq x_{3n+1} \geq 0 $$ Show that: $$ x_1^2 - x_2^2 + ... - x_{3n}^2 + x_{3n+1}^2 \geq (x_1 - x_2 + ... - x_{3n} + x_{3n+1})^2 $$ I'm trying to ...
3
votes
2answers
82 views

If I'm asked to prove that $n \le m$, is it sufficient to show that $n < m$?

I have a homework question, which is to prove by induction that $\sum\limits_{r=1}^{n} \frac{1}{\sqrt{r}} \leq 2\sqrt{n}$ for every integer $n \geq 1$. I've managed to show by induction that ...
0
votes
1answer
56 views

An inequality by induction

I am reading Arthur Engel's Problem Solving Strategies. Section 8. The Induction Principle Problem 24 with its solution are attached . I do not understand the second inequality in the solution on ...
2
votes
6answers
291 views

If $a_1,a_2,\ldots,a_n>0$ and $a_1+a_2+\ldots+a_n<\frac{1}{2}$, prove that $(1+a_1)(1+a_2)\ldots(1+a_n)<2$.

If $a_1,a_2,\ldots,a_n>0$ and $a_1+a_2+\cdots+a_n<\frac{1}{2}$, prove that $$(1+a_1)(1+a_2)\cdots(1+a_n)<2$$ I've tried using Holder's inequality (the same result can easily be derived using ...
0
votes
3answers
54 views

How to show $((k+1)!)^2 2^k \leq (2(k+1))!$

How do you show that $((k+1)!)^2 2^{k+1} \leq (2(k+1))!$ This is part of an induction proof and I have not made any progress.
0
votes
1answer
46 views

Prove the following inequality using induction: $(1 + \epsilon)^n \leq 1+ (2^n - 1)\epsilon$ for every $n \geq 1$ and $0 \leq \epsilon \leq 1$

Prove the following inequality using induction: $$(1 + \epsilon)^n \leq 1+ (2^n - 1)\epsilon$$ for every $n \in \mathbb{N}: n \geq 1$ and $0 \leq \epsilon \leq 1$ I'm familiar with the concept of ...
3
votes
2answers
299 views

Induction: show that $\sum\limits_{k=1}^n \frac{1}{\sqrt{k}} < 2 \sqrt{n}$ for all n $\in Z_+$

The question: show by using induction that $\sum\limits_{k=1}^n \frac{1}{\sqrt{k}} < 2 \sqrt{n}$ for all n $\in Z_+$ My attempt at a solution: The base case $n = 1$ is true. First we use the ...
2
votes
3answers
68 views

Prove $2^n\cdot n! ≤ (n+1)^n$ by induction.

An induction I'm struggling with. Prove $2^n\cdot n! ≤ (n+1)^n$ by induction. An idea was to show that $2^n\cdot n! ≤ 1+n^2$ since $1+n^2 ≤ (n+1)^n$ using Bernoulli. However the inequality is ...
1
vote
1answer
45 views

Proving $\prod \limits_{k=0}^{n}(1-a_k) \geq1- \sum\limits_{k=0}^{n}a_k$

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 ...
1
vote
4answers
156 views

Prove using induction : $n < 3^n$

$ p(n) = n < 3^n = q(n) $ when $n=1$, $p(n)=1< 3=q(n)$ Assume the result is true for $n=m$ $p(m)=m < 3^m$ when $n = m+1$ $p(m+1) = m+1 < 3^m +1<3*3^m = 3^{m+1}=q(m+1)$ is this ...
0
votes
4answers
66 views

Proving $\displaystyle\sum_{k=1}^{m+1} \frac{1}{\sqrt{k}}\gt\sqrt{m+1}$

well the original problem was to prove the sum of k to the negative one half was more that the square root of n but it thought it would be best to use induction and get the equation displayed above. I ...
-1
votes
2answers
50 views

Mathemathical Induction Inequality [closed]

I can't do this question please help me. Prove $3^n>n^2$ for $n\ge 2$ (and also for n=0 and 1)
0
votes
0answers
46 views

Inequality sine power series (induction)

How can we show, for $k\geq 1$ and $x \geq 0$, the inequality below by induction? $\displaystyle \sin x \geq \sum_{n=1}^{2k} (-1)^{n+1} \frac 1{ (2n - 1)! }x^{2n-1} $ The base case $k = 1$ gives ...
2
votes
0answers
73 views

Inequality sine power series

How can we show, for $k\geq 1$ and $x \geq 0$, the inequality below by induction? $\displaystyle \sin x \geq \sum_{n=1}^{2k} (-1)^{n+1} \frac 1{ (2n - 1)! }x^{2n-1} $ The base case $k = 1$ gives ...
4
votes
4answers
72 views

Proving a relation with induction

I have a problem: Let $p_n$ be the $n:th$ prime number ($p_1=2, p_2=3, p_3=5$ and so on). With induction, show that $p_{n+2}>3n$ for each integer $n\geq1$. I can't figure this out because the ...
3
votes
3answers
55 views

Prove inequality $n<3^n$ using mathematical induction

Prove that $n<3^n$ where $n \in \mathbb N$, when $ n=1 $, I have proved it's true. And assumed when $n = p$ , $p<3^p$ is true. Can any body help me in showing that it is true for $n =p+1$
3
votes
2answers
82 views

Is this how you prove by induction for inequalities?

the question is here: http://cpsc.ualr.edu/srini/DM/chapters/examples/ex2.3.2.html My solution is as below:
6
votes
3answers
140 views

Math Induction Proof: $(1+\frac1n)^n < n$

So I have to prove: For each natural number greater than or equal to 3, $$(1+\frac1n)^n<n$$ My work: Basis step: $n=3$ $$\left(1+\frac13\right)^3<3$$ $$\left(\frac43\right)^3<3$$ ...
3
votes
2answers
83 views

Prove that $\sum_{k=0}^n\frac{1}{k!}\geq \left(1+\frac{1}{n}\right)^n$ [duplicate]

It basically says it all in the title. I tried solving the inequality using the bernoulli inequality somehow $$\dfrac{\displaystyle\sum_{k=0}^n\frac{1}{k!}}{(1+\frac{1}{n})^n}\geq 1,$$ but the ...
2
votes
3answers
137 views

Induction: $\sqrt{2\sqrt{3\sqrt{\cdots\sqrt n}}} < 3$ [duplicate]

I'm trying to prove that $$ \sqrt{2\sqrt{3\sqrt{4\cdots\sqrt{n}}}} < 3 $$ for any $n$ and have decided to use strong induction and instead just show that $$ \sqrt{k\sqrt{(k + 1)\cdots\sqrt{n}}} ...
3
votes
3answers
78 views

How to prove the inequality $\sum_{i=1}^n \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{n}}{2} $ for $n\in\mathbb{Z}^+$?

I have to prove this inequality: $$ \forall n \in Z^+, \sum_{i=1}^n \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{n}}{2} $$ So far, I have done the base cases and assumed the inequality is true for some ...