Research infinite series concerns: in the end can not converge to a number? Convergence is essentially a series of
Constant Series
- Is essentially the number of columns, such as the study of {xn} is equal to the limit problem study sum {xn-xn-1} the series convergence problems
- Linear algorithms to meet: a bunch of convergence of linear operation after convergence is still
- For convergent series, the presence of the associative law (the number of stages after the formation of any new series bracketed still converges to the sum of a series) (Tip proof: monotone sequence converges which sub-sequence is convergent)
Necessary conditions: lim un -> 0
- At first glance this, if this is not met, it must not converge.
- That is a positive progression if single increase that must not converge (either limit does not exist or is not 0)
Important stages
- \begin{equation*} \sum ^{\infty }_{n=1} \ \frac{1}{n^{p}} \end{equation*}
- when p = 1 is
\ the begin {Gather *} \ SUM ^ {\ infty} _ {n-=. 1} \ FRAC {. 1} {\ \ n-} \ End {Gather *} , derivative lnx is 1 / x, so this series are similar to ln (n), and \ begin {gather *} \ lim _ {x \ rightarrow \ infty} \ \ \ \ ln x = \ infty \ end {gather *}, it is possible to know at which time the number of stages is not a number.
Then p <1 when it bigger so certainly is not a number
p> Convergence 1 (to be filled pit)
- \begin{gather*} \sum ^{\infty }_{n=0} \ a_{n} \ q^{n} \end{gather*}
Geometric sequence, q <1 converges
Positive series converges judge
Comparison of the Convergence
- Large income, low income must prove: monotone bounded theorem
- Small divergence, big divergence must prove: If a large convergence, the convergence on small contradictions
The Inspection
\begin{equation*} \lim _{n\rightarrow \infty } \ \ \frac{a_{n}}{a_{n-1}} \ < 1 \end{equation*}- With geometric series cards, there is always a limit common ratio> above, and then multiplied by a time equivalent to the common ratio, then the common ratio of the series converge, this is certainly smaller than its convergence
Square root
\begin{equation*} \lim _{n\rightarrow \infty } \ \ \sqrt[n]{a_{n}} \ < \ 1 \end{equation*}- Or a geometric sequence, after a power equivalent to the n number that is less than n-th power of a, with a provable larger than its limit of geometric progression q