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 ...
1
vote
1answer
71 views

Proof of an inequality involving factorials

How can the following inequality be proven? $$\left(n!\right)^{\frac{1}{n}}\left((n+1)!\right)^{-\frac{1}{n+1}}\gt\dfrac{n}{n+1}$$ I know this is a result obtained in 1964, but I don't know how to ...
8
votes
1answer
139 views

One-Line Proof for $n! \geq (\frac n e)^n$

I was told to find a one-line proof for $n! \geq (\frac n e)^n$. I'm advised that Stirling's formula is not helpful. I've spent a little bit of time on it, but the solution is not coming to me. I feel ...
5
votes
1answer
88 views

Solving an inequality : $n \geq 3$ , $n^{n} \lt (n!)^{2}$.

I proved this inequality in the following way: Lemma: $r \in \Bbb N, r \geq 3$. We have $r^r \gt (r+1)^{r-1}$. Proof: We apply the AM-GM inequality to the $r$ positive integers where there are ...
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 ...
8
votes
3answers
122 views

Showing $n!<e(\frac{n}{2})^n$

I'd like to prove that $n!<e(\frac{n}{2})^n$. What I have so far: $\sqrt[n]{n!} = \sqrt[n]{1\cdot 2 \cdot \ldots \cdot n} \leq \frac{1+\ldots +n}{n}=\frac{(n+1)n}{2n}=\frac{(n+1)}{2}$. Thus ...
10
votes
4answers
346 views

How to prove $\left(\frac{n}{n+1}\right)^{n+1}<\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}<\left(\frac{n}{n+1}\right)^n$

Show that: $$\left(\dfrac{n}{n+1}\right)^{n+1}<\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}<\left(\dfrac{n}{n+1}\right)^n$$ where $n\in \Bbb N^{+}.$ If this inequality can be proved, then we have ...
1
vote
4answers
149 views

Which is larger :: $y!$ or $x^y$, for numbers $x,y$.

This is a generalization of this question :: Which is larger? $20!$ or $2^{40}$?. No explicit general solution was presented there and I'm just curious :D Thank-you. Edit :: I want a most-general ...
10
votes
3answers
499 views

Which is greater, $300 !$ or $(300^{300})^\frac {1}{2}$?

Which is greater among $300 !$ and $\sqrt {300^{300}}$ ? The answer is $300 !$ (my textbook's answer). I do not know how to solve problems involving such large numbers.
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
61 views

How to solve the inequality $n! \le n^{n-2}$?

The inequality is $n! \le n^{n-2}$. I used Stirling's approximation for factorials and my answer was $n \le (e(2\pi)^{-1/2})^{2/5}$ but this doesn't seem right. Any help would be much appreciated.
2
votes
1answer
188 views

How do you solve an inequality with the factorial of a variable?

How do you solve an inequality with the factorial of a variable? Example: Determine the interval of $n \in \Bbb N$ for which the following inequality holds: $$n! \leq 157788 \cdot 10^{10} $$ Can ...
2
votes
2answers
79 views

Inequality $C\lceil\log{n}\rceil! \geq n^k$

I've been struggling to prove there exist $C$ for $n, n_{0}, \forall k >0 \in \mathbb{R}$ such that $\forall n > n_{0}$: \begin{equation}C\lceil\log{n}\rceil! \geq n^k\end{equation} As you ...
1
vote
0answers
41 views

What are the best and most elementary bounds for $n!$?

What this question is looking for is bounds on $n!$ that are elementary in nature (I seem to have a fetish for these type of proofs). In general, as the results become more complicated, they also ...
-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
6
votes
3answers
124 views

How to prove $(\frac{n+1}{e})^n<n!<e(\frac{n+1}{e})^{n+1}$ without integrating method?

How to prove $$\left(\frac{n+1}{e}\right)^n<n!<e\left(\frac{n+1}{e}\right)^{n+1}$$ without integrating method? In fact we could prove this by noticing that $$i<x<i+1\Rightarrow \ln ...
0
votes
1answer
48 views

Looking for suggestions on how to proceed with showing that:

for $x \ge 2863:$ $$\ln\left(\left\lfloor\frac{x}{6}\right\rfloor!\right) < \sum_{k=5}^{\infty}-\mu(k)\ln\left(\left\lfloor\frac{x}{k}\right\rfloor!\right)$$ I've written a java application which ...
0
votes
0answers
80 views

Using Stirling's Formula to approximate a difference of logarithms of factorials in the same way as Jitsuro Nagura.

In Jitsuro Nagura's classic proof of a prime existing between $x$ and $\frac{6x}{5}$, he uses Stirling's formula to show that: $$T\left(x\right) - T\left(\frac{x}{2}\right) - ...
0
votes
0answers
54 views

Do these inequalities regarding the gamma function and factorials work?

I am seeing if it is possible to generalize lemma 1 in Nagura's proof that there is always a prime between $x$ and $\frac{6x}{5}$. In a previous question, I asked whether the following inequality is ...
2
votes
0answers
60 views

Trying to generalize an inequality from Jitsuro Nagura: Does this work?

I am investigating the generality of Lemma 1 in Nagura's proof that there is always a prime between $x$ and $\frac{6x}{5}$. In Lemma 1, Nagura establishes when $n > 1$, $x \ge 1$: ...
0
votes
1answer
65 views

How do you evaluate an inequality that involves logarithms of factorials?

For $x > 1$, $n > 2$ with $2 \mid x+1$ and $n \mid x+1$, does it then follow that: $$\log(\lfloor\frac{x}{2}\rfloor!) - \log(\lfloor\frac{x}{n}\rfloor!) \le \log(\lfloor\frac{x+1}{2}\rfloor!) - ...
4
votes
0answers
136 views

Inequality problem with factorials

I am not sure if this kind of "question" is welcome on MSE. Here is an olympiad-like problem that I would like to share with you: Let $a,b,c$ be nonnegative integers. Prove that $$ ...
8
votes
1answer
226 views

Is it possible to generalize Ramanujan's lower bound for factorials that he used in his proof of Bertrand's Postulate?

I am starting to feel more confident in my understanding of Ramanujan's proof of Bertrand's postulate. I hope that I am not getting overconfident. In particular, Ramanujan's does the following ...
2
votes
2answers
213 views

How to prove this inequality with factorials

This is what I am trying to solve: $n!>n^{\frac {n}{2}}$ I tried with induction and somehow it doesn`t work this way, I tried with logarithms and also didnĀ“t find a way. Is there an elementary(if ...
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 ...
0
votes
1answer
41 views

Another inequality question

Can somebody in elementary way show that $(n!)^{2\over n+1}>n-1$ for only finitely many $n\in\mathbb N$? I need to prove this to be able to prove something else.
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 ...
5
votes
7answers
202 views

Some trouble with a proof on $n!/(\sqrt{n})^n \geq 1$

Originally the problem is to prove that $n! \geq n^{n/2}$. I reduced this to: $n! \geq (\sqrt{n})^n$ so that: Prove that $\frac{n!}{(\sqrt{n})^n} \geq 1$. Each term in $n!$ is divided by the ...
9
votes
4answers
295 views

Is $10^{8}!$ greater than $10^{10^9}$?

My question is: $10^8! > 10^{10^9}$ ? I know that factorial is greater than exponential, but I am not sure about this specific case. Thanks,
25
votes
6answers
1k views

$m!n! < (m+n)!$ Proof?

Prove that if $m$ and $n$ are positive integers then $m!n! < (m+n)!$ Given hint: $m!= 1\times 2\times 3\times\cdots\times m$ and $1<m+1, 2<m+2, \ldots , n<m+n$ It looks simple but ...
1
vote
2answers
209 views

Relating Gamma and factorial function for non-integer values.

We have $$\Gamma(n+1)=n!,\ \ \ \ \ \Gamma(n+2)=(n+1)!$$ for integers, so if $\Delta$ is some real value with $$0<\Delta<1,$$ then $$n!\ <\ \Gamma(n+1+\Delta)\ <\ (n+1)!,$$ because ...
1
vote
1answer
992 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 $n>2$.
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 ?
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} = ...