# sum of the first n

• Nov 19th 2007, 01:45 AM
perash
sum of the first n
If the sum of the first n terms in the sequence $a_1, a_2, a_3, ...$ is equal to 1/n, find the product of its first 2007 terms.
• Nov 19th 2007, 04:04 AM
Soroban
Hello, perash!

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

Quote:

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

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

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

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

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

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

. . Good luck!

• Nov 20th 2007, 08:40 AM
PaulRS
$\sum_{i=1}^n{a_i}=\frac{1}{n}$

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

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

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

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

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