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Alternating series
series whose terms are alternately positive and negative
Let a_n>0. The alternating series \sum_{n=1}^{\infty }(-1)^na_n and \sum_{n=1}^{\infty }(-1)^{n+1}a_n converge when …
\lim_{n \to \infty } a_n=0
a_{n+1}\le a_n for all values of n
How can I prove if a_{n+1}\le a_n for all values of n?
Way 1
let f(x) = inner part of series but not alternating series part
Find f’(x)
use f’(x) to prove if a_{n+1}\le a_n for all values of n
Way 2
Cross multiply a_{n+1} & a_n
simplify the terms
whichever simplified term is biggest is also the biggest regular term
the numerator used in the biggest simplified term is the numerator of the biggest term
What if \lim_{n \to \infty } a_n\neq0 ie the first condition fails
Use the n^{th} term test!
Remainder
Difference between the exact and approximate sum of the series.
R_n = S - S_n
aka error
If a series passes the alternating series test than the absolute value of the remainder is …
less than or equal to the first neglected term (ie first term not used in partial sum estimation)
\left| R_n \right|=\left| S-S_n \right|
\left| R_n \right|\le a_{n+1}
How to find a certain error
\left| R_n \right|\le a_{n+1}
R_n = the error
if the error needs to be less than a #, ex: c
a_{n+1}<c
Now isolate n
ex: # < n
so for an error less than c we need # +1 terms.
Absolute Convergence
If \sum_{}^{}\left| a_n \right| converges then \sum_{}^{}a_n also converges and the series is said to be absolutely convergent.
Conditionally Convergent
If \sum_{}^{}a_n converges but \sum_{}^{}\left| a_n \right| diverges the series is said to be conditionally convergent.