# Infinite Serie

• Nov 4th 2008, 09:28 AM
amiv4
Infinite Serie
If the nth partial sumof an infinite series a_n is s_n = 3 - n2^-n
find a_n and the infinite series a_n
• Nov 4th 2008, 11:49 AM
Soroban
Hello, amiv4!

Quote:

If the $n^{th}$ partial sum of an infinite series is: . $S_n \:= \:3 - n^2-n$

find $a_n$ . . . and the infinite series $a_n$ ?? .
The series diverges
Crank out the first few partial sums . . .

. . $\begin{array}{ccc} S_1 &=& 1 \\ S_2 &=& \text{-}3 \\ S_3 &=& \text{-}9 \\ S_4 &=& \text{-}17 \\ S_5 &-& \text{-}27 \\ \vdots &&\vdots \end{array}$

We can see that the summation is:

. . $S_n \;=\;1 - 4 - 6 - 8 - 10 - \hdots - 2n$

. $S_n \;=\;\overbrace{3 - 2} - 4 - 6 - 8 - \hdots - 2n$

. $S_n \;=\;3 -2(1 + 2 + 3 + 4 + \hdots + n)$

. $S_n \;=\;3 - 2\,\frac{n(n+1)}{2} \quad\hdots\quad\text{which equals: }3 - n^2-n\quad\hdots$ YAY!

I can't find a neat representation for $a_n$

. . $a_n \;=\;\bigg\{\begin{array}{ccc}1 & & n = 1 \\ \text{-}2n & & n > 1 \end{array}$