1
vote
4answers
66 views

Proving $4^n > n^4$ holds for $n\geq 5$ via induction.

I know that it holds for $n=5$, so the first step is done. For the second step, my IH is: $4^n > n^4$, and I must show that $4^{n+1} > (n+1)^4$. I did as follows: $4^{n+1} = 4*4^n > 4n^4$, ...
3
votes
3answers
48 views

Prove or Disprove Inequality By Induction

Prove or Disprove $\sum_{i=0}^n(2i)^3 \le (8n)^3 $ If true, prove using induction. If false, give the smallest value of n that is a counter example and the values for the left and right hand sides ...
3
votes
1answer
67 views

Proving that if $n>2$ then $n!>n^{n/2}$ using induction. [duplicate]

How to prove that if $n>2$ then $n!>n^{n/2}$ using induction?
1
vote
2answers
34 views

Proof an inequality by mathematical induction

I have a problem that I have to solve using mathematical induction but I'm stuck from a part. The problem is: Proof that $\large n<2^n$ is true for $\large n \in \mathbb{N}\ $ So, I did that ...
5
votes
5answers
212 views

Need to prove inequality $\sum\limits_{k=0}^n \frac{1}{(n+k)} \ge \frac{2}{3}$

Prove that for $n \geq 1$: $$\sum\limits_{k=0}^n \frac{1}{(n+k)} \ge \frac{2}{3}$$ I have tried math induction but that didn't work. Although I'm pretty sure that the solving can be done by induction ...
1
vote
1answer
63 views

Prove $1! + 2! + . . . + n! < (n + 1)!$ using mathematical induction [duplicate]

$1! + 2! + . . . + n! < (n + 1)!$ This question has left me stumped for quite some time. I am not sure how to approach it. (I am really bad at induction).
-1
votes
3answers
48 views

how to prove $1/n (1-(1/2)^n)$ decreasing without using differentiation

$a(n)=1/n (1-(1/2)^n)$ prove $a(n+1)<a(n)$ for n>0 by differentiating slope comes negative and then we can prove it . but i wanted to solve it without that . can someone help
2
votes
1answer
114 views

Use induction to prove the following: $1! + 2! + … + n! \le (n + 1)!$

Use induction to prove the following: $1! + 2! + .... + n! < (n + 1)!$ Base case: $n = 1$ $1! < 2!$ true Inductive step: Assume that $1! + 2! + .... + k! \le (k + 1)!$ is true let $n = k ...
2
votes
2answers
71 views

Show that $(n+1)^{n+1}>(n+2)^n$ for all positive integers

Show that: $(n+1)^{n+1}>(n+2)^n$ holds for all positive integers I tried using induction: for $n=1$ we have 4>3 then for $n+1$ we have to show that $(n+2)^{n+2}>(n+3)^{n+1}$ and here I ...
3
votes
3answers
85 views

Proof by induction: $n$th Fibonacci number is at most $ 2^n$

I'm trying to find the proof by induction of the following claim: For all $n\in\mathbb N$, $\operatorname{fibonacci}(n) \le 2^n$ My Proof: Base case: $n = 1$ $\operatorname{fibonacci}(1) \le 2^ 1$ ...
0
votes
2answers
54 views

Mathematical Induction - Inequality

Does anyone have any idea on how to complete the inductive step? Thm: For all $n >= 0~~~~ 6^n + 4 > n^3$ Pf: by Induction     Let $P(n)$ be proposition that $~6^n + 4 ...
1
vote
2answers
69 views

Induction inequality on sum of reciprocals

I have to prove that: $\displaystyle\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\ge\frac{1}{2}$ for natural $n$ Checking for $n=1$ we have $\displaystyle 1+\frac{1}{2}=\frac{3}{2}\ge \frac{1}{2}$ ...
4
votes
2answers
120 views

Prove that $\sqrt{n} \le \sum_{k=1}^n \frac{1}{\sqrt{k}} \le 2 \sqrt{n} - 1$ is true for $n \in \mathbb{N}^{\ge 1}$

I'm trying to solve these induction exercises proposed by the department of mathematics of Oxford University. I don't know how to give a valid proof for the third one which says the following: ...
1
vote
4answers
46 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 ...
2
votes
2answers
82 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
78 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
41 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
52 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
56 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
34 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
58 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
58 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
60 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
54 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
66 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
50 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
108 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
72 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
58 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
37 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
51 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
48 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
192 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
181 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
60 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
40 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
139 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
381 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
70 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
75 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
168 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
83 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
64 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
334 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 Hölder's inequality (the same result can easily be derived using ...
0
votes
3answers
56 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
48 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
350 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 ...