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- Thread starter MarkFL
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- Jan 26, 2012

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$$\sum_{k=1}^{j}\underbrace{\phantom{e^{-x^{2}}}}_{\text{What goes here?}}$$

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

- Jan 25, 2013

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$1-\dfrac {1}{1+2+3+----+n}=1-\dfrac {2}{n(n+1)}$Prove that:

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

I will use mathematical induction method to prove it

(it is easier)

j=2

$1-\dfrac{2}{6}=\dfrac {4}{6}$

j=3

$(1-\dfrac{2}{6})\times (1-\dfrac{2}{12})=\dfrac {5}{9}$

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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:

- Mar 31, 2013

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= (n(n+1) - 2)/(n(n+1) = (n+2)(n-1)/(n(n+1)

by multiplying out we get the result

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I also used induction...

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

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

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

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

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.