Tagged Questions
1
vote
0answers
99 views
Proving by induction that $n^{n+1} > (n+1)^n$ for $n \ge 3$ [duplicate]
Prove the following inequality by mathematical induction:
$$n^{n+1}>{(n+1)}^n \qquad (n\geq3)$$
Obviously it holds for $n=3$.
Assume $P(n)$ holds, then $LHS=n^{n+1}$
6
votes
1answer
69 views
Power tower inequality
I want to prove the following power tower inequality:
$$
3 \uparrow \uparrow 100 > 4 \uparrow \uparrow 99
$$
but I don't know how to do this. I think that induction will not work, because I think ...
8
votes
6answers
222 views
Show that $n \ge \sqrt{n+1}+\sqrt{n}$
(how) Can I show that:
$n \ge \sqrt{n+1}+\sqrt{n}$ ?
It should be true for all $n \ge 5$.
Tried it via induction:
$n=5$: $5 \ge \sqrt{5} + \sqrt{6} $ is true.
$n\implies n+1$:
I need to show ...
9
votes
3answers
364 views
Inductively prove that this sequence of integrals is bounded.
EDIT: I have an attempted solution to this in a post below, it is very long, but still incomplete.
EDIT:Alright, I've pretty much almost finished my solution, but my biggest problem is the 2nd ...
2
votes
3answers
59 views
problem with induction?
I am a bit new to logical induction, so I apologize if this question is a bit basic.
I tried proving this by induction:
$$\left(\sum_{k=1}^nk\right)^2\ge\sum_{k=1}^nk^2$$
Starting with the base ...
9
votes
2answers
127 views
Proving that $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{100}}<20$
How do I prove that:
$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{100}}<20$$
Do I use induction?
6
votes
5answers
86 views
Prove that $3^n>n^4$ if $n\geq8$
Proving that $3^n>n^4$ if $n\geq8$
I tried mathematical induction start from $n=8$ as the base case, but I'm stuck when I have to use the fact that the statement is true for $n=k$ to prove ...
2
votes
1answer
49 views
Prove that $n^2 \leq \frac{c^n+c^{-n}-2}{c+c^{-1}-2}$.
Let $c \not= 1$ be a real positive number, and let $n$ be a positive integer.
Prove that $$n^2 \leq \frac{c^n+c^{-n}-2}{c+c^{-1}-2}.$$
My initial thought was to try and induct on $n$, but the ...
4
votes
4answers
95 views
Prove that for every positive integer $n$, $1/1^2+1/2^2+1/3^2+\cdots+1/n^2\le2-1/n$
Base case: n=1. $1/1\le 2-1/1$. So the base case holds.
Let $n=k\ge1$ and assume
$$1/1^2+1/2^2+1/3^2+\cdots+1/k^2\le 2-1/k$$
We want to prove this for $k+1$, i.e.
...
2
votes
1answer
82 views
2
votes
3answers
124 views
Showing that $3n<n!$ whenever $n$ is an integer with $n \geq 7$
How can we show that:
$$3n< n!$$
whenever $n$ is an integer such that $n \geq 7$ ?
I was thinking that we can prove this by showing that such case is true with any integer above 7, but ...
4
votes
2answers
95 views
Prove $((n+1)!)^n < 2!\cdot4!\cdots(2n)!$
so I know I need to prove this via induction, but I am somewhat stuck. Here is what I have does so far.
Let $p(n) = (n+1)!^n \le 2!\cdot4!\cdot\ldots\cdot(2n)!$
$p(2) = 3!^2\le 2!\cdot4!$
Assume ...
2
votes
1answer
40 views
Unable to prove inequality using induction or understand intuitively
While evaluating the complexity of an algorithm, I got an inequality that I am unable to understand.
Let us consider this series:
$$x= n + (2^0 + 2^1 + 2^2 + ....... + ...
1
vote
3answers
67 views
Proving $(5-\frac5k )(1+\frac{1}{(k+1)^2}) \le 5 - \frac{5}{k+1}$
Can anyone tell me what I am doing wrong? need to prove for $k\ge2$
$$(5-\frac5k )(1+\frac{1}{(k+1)^2}) \le 5 - \frac{5}{k+1}$$$$(5-\frac5k )(1+\frac{1}{(k+1)^2})= 5(1-\frac1k)(1+\frac1{(k+1)^2})$$
...
4
votes
2answers
83 views
Demonstrate by induction the inequality: $\ln(1+n)\leq\sum_{i=1}^{n}\frac{1}{i}\leq1+\ln(n)$
Kind of stuck in this one. I've tried substracting
$$\ln(1+(n+1))\leq\sum_{i=1}^{n}\frac{1}{i}+\frac{1}{n+1}\leq1+\ln(n+1)$$
at the original inequality and applying properties of the logarithms, but ...
0
votes
0answers
51 views
proving an inequality by induction
Not sure how to proceed. I'm trying to prove that the following inequality is true. I know that $t_2 = 6$ and $t_3=17$ from the problem statement. The base case is obvious.
$t_{r+1} \leq (r+1) (t_r ...
5
votes
1answer
61 views
Proving the following inequality using induction?
I need to prove by induction that $Q_n \ge\frac{10}{3}-\frac5{3n}$ for $n\ge2$$$
Q_n = ...
7
votes
1answer
121 views
An inequality using mathematical induction
It was shown in here that $\left(1+\frac{1}{n}\right)^n < n$ for $n>3$. I think we can be come up with a better bound, as follows:
$$\left(1+\frac{1}{n}\right)^n \le 3-\frac{1}{n}$$
for ...
1
vote
2answers
57 views
Non-induction proof of $2\sqrt{n+1}-2<\sum_{k=1}^{n}{\frac{1}{\sqrt{k}}}<2\sqrt{n}-1$
Prove that $$2\sqrt{n+1}-2<\sum_{k=1}^{n}{\frac{1}{\sqrt{k}}}<2\sqrt{n}-1.$$
After playing around with the sum, I couldn't get anywhere so I proved inequalities by induction. I'm however ...
5
votes
4answers
122 views
Proving ${(n+2)!}< \left(n(\sqrt{2}-1)+\sqrt{2}\right)^{n+2}$
While simplifying an inequality, this inequality was derived:
$${(n+2)!}< \left(n(\sqrt{2}-1)+\sqrt{2}\right)^{n+2},\quad\quad\quad\quad n\in \mathbb{N}$$
Do you have any idea to prove it? It is ...
4
votes
4answers
143 views
Help in proving $ \left( 1 + \frac{1}{n} \right)^{n} \leq \sum\limits_{k=0}^{n} \frac{1}{k!} < 3 $.
I am trying to prove this statement for all $ n \geq 1 $ using induction:
$$
\left( 1 + \frac{1}{n} \right)^{n} \leq \sum_{k=0}^{n} \frac{1}{k!} < 3.
$$
I said:
Base case $ n = 1 $:
$$
\left( ...
6
votes
3answers
113 views
Proof of inequality using induction
Prove that $2(\sqrt{n+1}-\sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n} - \sqrt{n-1})$ if $n \ge 1$ using induction.
Can someone help me with this problem please. Base case is easily shown, and ...
0
votes
2answers
121 views
4
votes
1answer
140 views
Please prove that ${n\choose r} < (n+1)^r$. My induction-based proof is ugly!
I came across this inequality in the University of Missouri Youtube Channel lecture 38:
College Algebra - Lecture 38 - The Binominal Theorem
The professor asks us 1:01:00 into the video to prove ...
0
votes
1answer
72 views
Why this inequality yields at most exponential growth?
Let $\Omega=\mathbf{R}^{n-j}\times\omega$, where $\omega\subset\mathbf{R}^j$ is a smooth bounded domain. Consider a function $u:\overline\Omega\rightarrow\mathbf{R}$ that satisfies
$$u(x,y)+k\leq ...
1
vote
2answers
136 views
Proof by induction - Being stuck before being able to prove anything!
I am currently in the process of learning how to prove statements
by induction. For the most part, I understand but I am at most
clueless when it comes to prove some inequalities.
For example, I am ...
1
vote
2answers
60 views
Prove, formally that: $\log_2 n! \ge n$ , for all integers $n>3$.
Prove, formally that:
$\log_2 n! \ge n$ for all integers $n>3$.
Hint: first prove that $n! ≥2^n$, for all integers $n >3$.
So far what I have:
Base case, $n = 4$,
$4! = 24$
$2^4 = 16$.
...
2
votes
1answer
78 views
Is this induction procedure correct? ($2^n<n!$)
I am rather new to mathematical induction. Specially inequalities, as seen here How to use mathematical induction with inequalities?. Thanks to that question, I've been able to solve some of the form ...
1
vote
2answers
57 views
Mathematics Induction on Inequality
I want to prove
$2^n \ge 3n^2 +5$--call this statement $S(n)$--for $n\ge8$
Basis step with $n = 8$, which $\text{LHS} \ge \text{RHS}$, and $S(8)$ is true. Then I proceed to inductive step by ...
11
votes
8answers
482 views
prove $\frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}<\frac{25}{36}$ by mathematical induction
How to prove
$$\frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}<\frac{25}{36}$$
by Mathematical induction，n$\ge $1
0
votes
1answer
30 views
correctness of inequality
Let $k \geq 6$ and I know $k!$ < $\dfrac{k^k}{2^k}$ I want to show the following:
$(k+1)! < \dfrac{(k+1)^{k+1}}{2^{k+1}}$
Now I am going to show my solution, let me know if my reasoning is ...
3
votes
1answer
139 views
inequality with sum of powers
How to prove the following inequality:
$$\forall n\geqslant 4:\dfrac {3^{n}+4^{n}+\cdots +\left( n+2\right) ^{n}} {\left( n+3\right) ^{n}} < 1$$
1
vote
4answers
676 views
Proving inequality by induction
I am trying to prove for all natural $n$ that:
$$5^n + 5 < 5^{n+1}$$
I did the basic step with $n=1$ and inequality holds, I am now at the induction step:
$$5^{k+1} + 5 < 5^{k+2}$$
and I have ...
2
votes
5answers
791 views
Prove by mathematical induction that $2n ≤ 2^n$, for all integer $n≥1$?
I need to prove $2n \leq 2^n$, for all integer $n≥1$ by mathematical induction?
This is how I prove this:
Prove:$2n ≤ 2^n$, for all integer $n≥1$
Proof: $2+4+6+...+2n=2^n$
$i.)$ Let $P(n)=1
...
3
votes
2answers
158 views
Lower Bound of Central Binomial Coefficients
I would like to prove by induction the following inequality:
$\frac{4^n}{n+1} < \binom{2n}{n}$, for all natural numbers n > 1.
Any hints?
4
votes
1answer
79 views
Is my proof for $\sum_{i=1}^{n}x_{i}y_{i}\leq \sqrt{\sum_{i=1}^{n}x_{i}^{2}\sum_{i=1}^{n}y_{i}^{2}}$ correct?
Is this part of my proof by induction correct ?
$\sum_{i=1}^{n}x_{i}y_{i}\leq \sqrt{\sum_{i=1}^{n}x_{i}^{2}\sum_{i=1}^{n}y_{i}^{2}}$
this is true when the true is that :
$\sum_{i=1}^{n}\left ...
1
vote
1answer
253 views
Proof by induction of triangle inequality in Hilbert space.
I've made proof by induction over $n$ for triangle inequality : $\left \| x+y \right \|_{e}\leq \left \| x \right \|_{e}+\left \| y \right \|_{e}$
,where $\left \| x \right ...
1
vote
1answer
427 views
Prove by induction that $n^3 > n^2 − 6n + 4$ for all $n ∈ {\mathbb N}$ with $n ≥ 1$ .
Would you please check if my answer is correct and confirm that it is proof by induction? Thank you.
The proof is by induction.
Base Case: when $n=1$:
...
1
vote
1answer
227 views
Assuming inequalities to be equal when proving by induction
I was struggling to get my head around proof by induction for inequalities when I came across a method described at The Math Forum (first answer). Here Dr. Ian goes ahead and compares the changes in ...
5
votes
4answers
460 views
Proof by induction for inequalities.
I have been trying to do this problem by using induction but I became stuck halfway through.
The problem:
Use induction to show that
$$2\left(1+ \frac{1}{8} + \frac{1}{27} + \cdots + ...
4
votes
2answers
153 views
Proving $ \left(\frac{n^2+1}{n^2}\right)^n\ge\frac{n+1}{n}$ by induction.
Prove for $n\in\mathbb{N}$:
$$ \left(\frac{n^2+1}{n^2}\right)^n\ge\frac{n+1}{n}.$$
by induction.
I'm doing induction ahead of my regular classes because I need it for competition coming in ...
7
votes
7answers
317 views
Please help me prove by induction that $n^n>1\cdot3\cdot5\cdot\ldots\cdot(2n-1)$
Please help me prove by induction that
$$
n^n>1\cdot3\cdot5\cdot\ldots\cdot(2n-1)
$$
1
vote
0answers
115 views
Prove the inequality $n! > 2^n$ by induction. [duplicate]
Possible Duplicate:
Proof the inequality $n! \geq 2^n$ by induction
Prove By Induction that $n!>2^n$
I have to prove the inequality $n! > 2^n$ for all integers $n \geq4$.
I am ...
4
votes
1answer
818 views
Proving the Schwarz Inequality for Complex Numbers using Induction
I want to prove the following version of the Schwarz Inequality for complex numbers $a_1, a_2, \ldots, a_n \in \mathbb{C}$ and $b_1, b_2, \ldots, b_n \in \mathbb{C}$:
$$|\sum_{j=1}^n a_j ...
1
vote
1answer
370 views
Inductive proof of Cauchy's inequality for complex numbers?
I'm trying to put together an inductive proof of Cauchy's inequality for the complex case,
$$
\left|\sum_{i=1}^na_ib_i\right|^2\leq\sum_{i=1}^n|a_i|^2\sum_{i=1}^n|b_i|^2.
$$
The base case is easy, ...
4
votes
2answers
423 views
How do I prove $(1 + \frac{1}{n})^n < n$ by mathematical induction?
$\displaystyle(1 + \frac{1}{n})^n < n$ for $n \gneq 3$
yes for $n = 1$ it is true
I assume it is true for $n = k$ and get
$\displaystyle(1+\frac{1}{k})^k < k$
I then go to $\displaystyle(1 ...
9
votes
10answers
273 views
Prove by induction that for all $n \geq 3$: $n^{n+1} > (n+1)^n$
I am currently helping a friend of mine with his preperations for his next exam. A big topic of the exam will be induction, thus I told him he should practice this a lot. As at the beginning he had no ...
2
votes
3answers
223 views
Strong Mathematical Induction Recursion Inequality
I have a question that is for a homework assignment and I just would like to ask if I seem to be on the right track or if I'm just doing it completely wrong.
Here is the question:
The sequence ...
1
vote
1answer
435 views
Proof the inequality $n! \geq 2^n$ by induction
I'm having difficulity solving an exercise in my course.
The question is:
Prove that $n!\geq 2^n$.
We have to do this with induction. I started like this:
The lowest natural number where the ...
5
votes
5answers
381 views
Prove an inequality by Induction: $(1-x)^n + (1+x)^n < 2^n$
Could you give me some hints, please, to the following problem.
Given $x \in \mathbb{R}$ such that $|x| < 1$. Prove by induction the following inequality for all $n \geq 2$:
$$(1-x)^n + (1+x)^n ...
