# Thread: sum of the first n

1. ## sum of the first n

If the sum of the first n terms in the sequence$\displaystyle a_1, a_2, a_3, ...$ is equal to 1/n, find the product of its first 2007 terms.

2. Hello, perash!

I must be missing some simple. .I'm getting no pattern . . .

If the sum of the first $\displaystyle n$ terms in the sequence: $\displaystyle a_1,\:a_2,\:a_3,\:\cdots$ is equal to $\displaystyle \frac{1}{n}$,
find the product of its first 2007 terms.

We are told that: .$\displaystyle \begin{array}{ccc}a_1 & = & 1 \\ a_1+a_2 & = & \quad\frac{1}{2} \\ a_1+a_2+a_3 & = & \frac{1}{3} \\ a_1+a_2+a_3+a_4 &=& \quad\frac{1}{4} \\ \vdots & & \vdots\end{array}$

This gives us: .$\displaystyle \begin{array}{ccccc} a_1 &=& 1 & = & \frac{1}{1!} \\ a_2 &=& \quad\text{-}\frac{1}{2} &=& \quad\text{-}\frac{1}{2!} \\ a_3 &=&\frac{5}{6} &=& \frac{5}{3!} \\ a_4 &=& \quad\text{-}\frac{7}{12} &=&\quad\text{-}\frac{14}{4!} \\ a_5 &=& \frac{47}{60} &=&\frac{94}{5!} \\ a_6 &=& \quad\text{-}\frac{37}{60} &=& \quad\text{-}\frac{444}{6!} \\ \vdots & & & & \vdots \end{array}$

The denominators are: .$\displaystyle 1!,\:2!,\:3!,\,\cdots\:2007!$

The numerators are: .$\displaystyle 1,\:\text{-}1,\:5,\:\text{-}14,\:94,\:\text{-}444,\:\cdots$
. . and I see no pattern there . . .

The desired product is: .$\displaystyle P_{2007} \;=\;\frac{(1)(\text{-}1)(5)(\text{-}14)(94)(\text{-}444)\:\cdots} {(1!)(2!)(3!)(4!) \cdots(2007!)}$

. . Good luck!

3. $\displaystyle \sum_{i=1}^n{a_i}=\frac{1}{n}$

But $\displaystyle \sum_{i=1}^{n-1}{a_i}=\frac{1}{n-1}$

Then: $\displaystyle a_n=\frac{1}{n}-\frac{1}{n-1}=-\frac{1}{n\cdot{(n-1)}}$ for $\displaystyle n\geq{2}$

So: $\displaystyle P_n=\prod_{i=1}^{n}{a_i}=(-1)^{n-1}\prod_{i=2}^{n}{\frac{1}{i\cdot{(i-1)}}}$

$\displaystyle P_n=(-1)^{n-1}\prod_{i=2}^{n}{\frac{1}{i\cdot{(i-1)}}}=(-1)^{n-1}\prod_{i=2}^{n}{\frac{1}{i}}\prod_{i=1}^{n-1}{\frac{1}{i}}$

Therefore: $\displaystyle P_n=(-1)^{n-1}\cdot{\frac{1}{n!\cdot{(n-1)!}}}$ when $\displaystyle n\geq{2}$