1
vote
1answer
57 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
77 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
48 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
44 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
374 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
951 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
82 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
84 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
53 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
97 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
273 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
148 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
109 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 ...
0
votes
2answers
135 views

Prove that $n! ≥ (⌈n/2⌉)^{⌈n/2⌉}$

Prove that : $n! ≥ (⌈n/2⌉)^{⌈n/2⌉}$
0
votes
2answers
144 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
112 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 ...
6
votes
1answer
649 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
897 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
370 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 ...
9
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
340 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
933 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} = ...