MHB Question: What is the value of ⌊ 2020/(1+2+3+...+2019)⌋?

  • Thread starter Thread starter juantheron
  • Start date Start date
juantheron
Messages
243
Reaction score
1
Finding value of $\displaystyle \bigg\lfloor \frac{2020!}{1!+2!+3!+\cdots +2019!}\bigg\rfloor$
 
Mathematics news on Phys.org
My attempt:
For $n > 2$ the following is true:

\[F_n = \left \lfloor \frac{(n+1)!}{1!+2!+3!+...+n!} \right \rfloor = n-1\]

Proof by induction:
Base cases:
\[F_3 =\left \lfloor \frac{4!}{1!+2!+3!} \right \rfloor = \left \lfloor \frac{24}{9} \right \rfloor = 2. \\\\ F_4 = \left \lfloor \frac{5!}{1!+2!+3!+4!} \right \rfloor = \left \lfloor \frac{120}{33} \right \rfloor = 3.\]

Suppose the identity holds for some n = m > 4. We need to show, that the identity also holds for n = m+1.

We have the identity: \[ F_m =\left \lfloor \frac{(m+1)!}{1!+2!+3!+...+m!} \right \rfloor = m-1.\]

To ease the algebra, let \[\sigma = 1!+2!+3!+...+m!\]

Then, we can write:

\[F_{m+1}=\left \lfloor \frac{(m+2)!}{1!+2!+3!+...+(m+1)!} \right \rfloor \\\\ =\left \lfloor \frac{(m+1)!}{\sigma +(m+1)!}(m+2) \right \rfloor\\\\ =\left \lfloor \frac{\frac{(m+1)!}{\sigma }}{1+\frac{(m+1)!}{\sigma }}(m+2) \right \rfloor\]

We know, that \[\frac{(m+1)!}{\sigma } = m-1+\varepsilon\] for some $0< \varepsilon<1$.

In other words: \[F_{m+1} =\left \lfloor \frac{m-1+\varepsilon }{m + \varepsilon }(m+2) \right \rfloor =\left \lfloor \left ( 1-\frac{1}{m+\varepsilon } \right )(m+2) \right \rfloor\]

Now, the fraction $\frac{m+2}{m+\varepsilon}$ has the sharp limits: \[1 < \frac{m+2}{m+\varepsilon } <2\]

This follows from the inequalities: $\varepsilon < 2 < m +2\varepsilon$

Thus the fraction can be written as: $\frac{m+2}{m+\varepsilon} = 1+\delta$ for some $0 < \delta < 1$.
Finally, we get

\[F_{m+1} =\left \lfloor m+2- (1+\delta )\right \rfloor = \left \lfloor m \right \rfloor+\left \lfloor 1-\delta \right \rfloor = m.\] q.e.d.- and we conclude, that $F_{2019}= 2018.$
 
Thanks Ifdahl for nice solution

Here is mine

[sp]Using $n!=n(n-1)! = [(n-1)+1](n-1)! = (n-1)(n-1)!+(n-1)(n-2)!$

So $(n-1)(n-1)!+(n-1)(n-2)!<(n-1)(n-1)!+(n-1)(n-2)!+(n-1)(n-3)!+\cdots (n-1)1!\;\forall n\geq 4$

So $n!<(n-1)\bigg[(n-1)!+(n-2)!+(n-3)!+\cdots +2!+1!\bigg]\cdots \cdots (1)$

And $n!=n(n-1)!=[(n-2)+2](n-1)!=(n-2)(n-1)!+2(n-1)!=(n-2)(n-1)!+2(n-1)(n-2)!$

So $n!=(n-2)(n-1)!+(n-2)(n-2)!+n(n-2)(n-3)!$

As $n(n-3)!>(n-3)!+(n-4)!+\cdots +2!+1!\forall n\geq 4$

So $n(n-2)(n-3)!>(n-2)\bigg[(n-3)!+(n-4)!+\cdots +2!+1!\bigg]$

So $n!>(n-2)\bigg[(n-1)!+(n-2)!+\cdots\cdots +2!+1!\bigg]\cdots \cdots (2)$

From $(1)$ and $(2),$ We have

$\displaystyle (n-2)<\frac{n!}{1!+2!+3!+\cdots \cdots +(n-1)!}<(n-1)$

So we get $\displaystyle \bigg\lfloor \frac{n!}{1!+2!+3!+\cdots \cdots +(n-1)!}\bigg \rfloor =(n-2)$

Now put $n=2010,$ We get $\displaystyle \bigg\lfloor \frac{2020!}{1!+2!+3!+\cdots \cdots +2019!}\bigg \rfloor =2018$[/sp]
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top