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 ...
0
votes
3answers
44 views

Proving by induction that $(n^2)!>(n!)^2$ for $n \geq 2$

I'm trying to prove that $(n^2)!>(n!)^2$ for $n \in [2,\infty) \cap\mathbb{Z^+}.$ Ok, here's what I've tried: $n \geq 2,$ $(n^2)!>(n!)^2$ ...
10
votes
3answers
138 views

$1!+2!+\ldots+n!$ cannot be the square of a positive integer

I have to prove that $1!+2!+\ldots+n!$ cannot be the square of a positive integer, $\forall n\geq4$. I've tried to do this with induction, but I don't seem to reach any satisfactory conclusion. Any ...
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 ...
1
vote
1answer
58 views

How does $(k+1)!(k+2)-1 = (k+2)!-1$?

I'm trying to do a proof by induction question and I'm at the very last part. Apparently $(k+1)!(k+2)-1 = (k+2)!-1$. I have checked using an online calculator. I don't understand why though.
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
1answer
57 views

Proving a sequence with induction reasoning

I have an assignment which I am quite stuck on. The question is the following: function f: N to N is defined recursivly: ...
0
votes
1answer
46 views

help on manipulating this algebraic expression

So I have something like: $\frac {k!}{(k-3)!3!}$ I'm going to add $\frac 12k(k-1)$ to this, and I want to obtain $\frac {(k+1)!}{(k-2)!3!}$ as the result. I'm having trouble with this since I need ...
1
vote
1answer
541 views

Proof by induction Involving Factorials

My "factorial" abilities are a slightly rusty and although I know of a few simplifications such as: $(n+1)\,n! = (n+1)!$, I'm stuck I have to prove by induction that: $$\sum_{i=1}^n\frac{i-1}{i!} = ...
0
votes
1answer
1k views

Mathematical Induction Factorials, sum r(r!) =(n+1)! -1 [duplicate]

How do I prove that $$\sum\limits_{r=1}^{n} r(r!) = (n+1)!-1$$ I was able to get to factor: $LHS = k(k!) + (k+1)(k+1)!$ $\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\, RHS = (k+2)! -1$
1
vote
1answer
100 views

Trying to understand an exercise using factorials with induction

Exercise: Prove that (n + 1)! - n! = n(n!) for any n $\ge$ 1 Given Answer: I will skip the basic step since I understand that part. (n + 2)! - (n + 1)! = (n + 1)!(n + 2) - n!(n + 1) I understand ...
1
vote
2answers
89 views

Prove $ \sum_{1\leq k < n} k^{\underline{m}}=\frac{n^{\underline{m+1}}}{m+1} $ by induction on $m$

I want to prove by induction the following sum: $$ \sum_{1\leq k < n} k^{\underline{m}}=\frac{n^{\underline{m+1}}}{m+1} $$ but induction should be on $m$. Any hint will be helpful. EDIT: $ ...
4
votes
3answers
61 views

Proof by induction that $\sum_{i=1}^n \frac{i}{(i+1)!}=1- \frac{1}{(n+1)!}$

Prove via induction that$\sum_{i=1}^n \frac{i}{(i+1)!}=1- \frac{1}{(n+1)!}$ Having a very difficult time with this proof, have done pages of work but I keep ending up with 1/(k+2). Not sure when to ...
-1
votes
3answers
104 views

Prove that $n! \geq 2^{n-1}$ for $ n\geq1$ [duplicate]

Mathematical Induction:-Prove that $n! \geq 2^{(n-1)}$ for $n\geq 1$. I tried mathematical induction but could not
4
votes
3answers
297 views

Prove that $(a+1)(a+2)…(a+b)$ is divisible by $b!$ [duplicate]

The problem is following, prove that: $$(a+1)(a+2)...(a+b)\text{ is divisible by } b!\text{ for every positive integer a,b}$$ I've tried solving this problem using mathematical induction, but I ...
2
votes
3answers
152 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 ...
5
votes
2answers
111 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 ...
1
vote
2answers
157 views

Prove that $n! ≥ (⌈n/2⌉)^{⌈n/2⌉}$ [closed]

Prove that : $n! ≥ (⌈n/2⌉)^{⌈n/2⌉}$
0
votes
2answers
148 views

Prove by induction that $n! > n^2$

How does one prove by induction that $n! > n^2$ for $n \geq 4$
2
votes
1answer
118 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 ...
7
votes
1answer
689 views

How to prove that $\mathrm{Fibonacci}(n) \leq n!$, for $n\geq 0$

I am trying to prove it by induction, but I'm stuck $$\mathrm{fib}(0) = 0 < 0! = 1;$$ $$\mathrm{fib}(1) = 1 = 1! = 1;$$ Base case n = 2, $$\mathrm{fib}(2) = 1 < 2! = 2;$$ Inductive case ...
1
vote
4answers
145 views

False statement proven by induction?: $ n\geq a\Rightarrow n!\geq a^n, n\in \mathbb{N}-\left \{ 0 \right \}$

Can you spot my mistake? I will show the false statement, that $n\geq a\Rightarrow n!\geq a^n, n\in \mathbb{N}-\left \{ 0 \right \}$, with induction For $n=1$ , $1\geq a\Rightarrow 1!\geq ...
1
vote
1answer
990 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 ...
2
votes
2answers
386 views

How to prove $n < n!$ if $n > 2$ by induction?

I am stuck with the question below, Prove by mathematical induction that $n<n!$ for $n>2$.
6
votes
3answers
1k views

Inductive proof for the Binomial Theorem for rising factorials

I want to proove the following equality containing rising factorials $$(x+y)^\overline{n}\overset{(*)}{=}\sum_{k=0}^n\binom{n}{k}x^\overline{k}y^\overline{n-k}.$$ For $n=1$ this equality is ...
10
votes
4answers
2k views

Solve by induction: $n!>(n/e)^n$

To Prove : $n! > (n/e)^n$ The question seems easy but it ain't; anyone up for it ?
2
votes
1answer
363 views

Use of algebra and factorials for a question related to proof by induction

$$ \begin{align*} &= (n+1)! − 1 + ( (n+1) · (n+1)! )\\ &= (n+1)! (1+n+1) − 1\\ &= (n+1)! (n+2) − 1\\ &= (n+2)! − 1\\ \end{align*} $$ I'm confused at how the first ...
4
votes
2answers
962 views

How to prove $a^n < n!$ for all $n$ sufficiently large, and $n! \leq n^n$ for all $n$, by induction?

I have a couple things I want to prove. I'm pretty sure a proof by induction is the best route for these. First, I need to show that $5^n < n!$ from some $n_{0} > 0$. I'm choosing $n_{0} = ...