Welcome to our community

Be a part of something great, join today!

Products and Sums and Proofs...oh, my!

  • Thread starter
  • Admin
  • #1

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Prove that:

\(\displaystyle \prod_{j=2}^n\left(1-\frac{1}{\sum\limits_{k=1}^j(k)} \right)=\frac{n+2}{3n}\)
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
Re: products and sums and proofs...oh, my!

$$\sum_{k=1}^{j}\underbrace{\phantom{e^{-x^{2}}}}_{\text{What goes here?}}$$
 
  • Thread starter
  • Admin
  • #3

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: products and sums and proofs...oh, my!

$$\sum_{k=1}^{j}\underbrace{\phantom{e^{-x^{2}}}}_{\text{What goes here?}}$$
I have fixed my OP to include that important piece of information. Thank you for catching this error. :D
 

Albert

Well-known member
Jan 25, 2013
1,225
Prove that:

\(\displaystyle \prod_{j=2}^n\left(1-\frac{1}{\sum\limits_{k=1}^j(k)} \right)=\frac{n+2}{3n}\)
$1-\dfrac {1}{1+2+3+----+n}=1-\dfrac {2}{n(n+1)}$
I will use mathematical induction method to prove it
(it is easier)
j=2
$1-\dfrac{2}{6}=\dfrac {4}{6}$
j=3
$(1-\dfrac{2}{6})\times (1-\dfrac{2}{12})=\dfrac {5}{9}$
--------
--------
suppose j=n-1
$(1-\dfrac{2}{6})\times (1-\dfrac{2}{12})\times ------\times (1-\dfrac {2}{(n-1)n})=\dfrac {n+1}{3(n-1)}$
then j=n
$(1-\dfrac{2}{6})\times (1-\dfrac{2}{12})\times --\times (1-\dfrac {2}{(n-1)n})\times (1-\dfrac {2}{n(n+1)})$
$=\dfrac {n+1}{3(n-1)}\times (1-\dfrac {2}{n(n+1)})=\dfrac {n+2}{3n}$
so the proof is completed
 
Last edited:

kaliprasad

Well-known member
Mar 31, 2013
1,309
1−1/(1+2+3+−−−−+n)=1−2/(n(n+1))
= (n(n+1) - 2)/(n(n+1) = (n+2)(n-1)/(n(n+1)

by multiplying out we get the result
 
  • Thread starter
  • Admin
  • #6

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I wish to thank everyone that participated! (Yes)

I also used induction...

First, show the base case $P_2$ is true:

\(\displaystyle \prod_{j=2}^2\left(1-\frac{1}{\sum\limits_{k=1}^j(k)} \right)=\frac{2+2}{3(2)}\)

\(\displaystyle 1-\frac{1}{\sum\limits_{k=1}^2(k)}=\frac{4}{6}\)

\(\displaystyle 1-\frac{1}{1+2}=\frac{2}{3}\)

\(\displaystyle \frac{2}{3}=\frac{2}{3}\)

Thus, the base case is true. Next, state the induction hypothesis $P_n$:

\(\displaystyle \prod_{j=2}^n\left(1-\frac{1}{\sum\limits_{k=1}^j(k)} \right)=\frac{n+2}{3n}\)

As the inductive step, multiply by:

\(\displaystyle 1-\frac{1}{\sum\limits_{k=1}^{n+1}(k)}=\frac{n(n+3)}{(n+1)(n+2)}\)

and we have:

\(\displaystyle \prod_{j=2}^{n+1}\left(1-\frac{1}{\sum\limits_{k=1}^j(k)} \right)=\frac{n+2}{3n}\left(\frac{n(n+3)}{(n+1)(n+2)} \right)\)

\(\displaystyle \prod_{j=2}^{n+1}\left(1-\frac{1}{\sum\limits_{k=1}^j(k)} \right)=\frac{1}{3n}\left(\frac{n(n+3)}{n+1} \right)\)

\(\displaystyle \prod_{j=2}^{n+1}\left(1-\frac{1}{\sum\limits_{k=1}^j(k)} \right)=\frac{(n+1)+2}{3(n+1)}\)

We have derived $P_{n+1}$ from $P_n$ thereby completing the proof by induction.