Convergent series

Clash Royale CLAN TAG#URR8PPP In mathematics, a series is the sum of the terms of an infinite sequence of numbers.
Given an infinite sequence (a1, a2, a3,…)displaystyle left(a_1, a_2, a_3,dots right), the nth partial sum Sndisplaystyle S_n
is the sum of the first n terms of the sequence, that is,
- Sn=∑k=1nak.displaystyle S_n=sum _k=1^na_k.
A series is convergent if the sequence of its partial sums S1, S2, S3,…displaystyle leftS_1, S_2, S_3,dots right tends to a limit; that means that the partial sums become closer and closer to a given number when the number of their terms increases. More precisely, a series converges, if there exists a number ℓdisplaystyle ell
such that for any arbitrarily small positive number εdisplaystyle varepsilon
, there is a (sufficiently large) integer Ndisplaystyle N
such that for all n≥ Ndisplaystyle ngeq N
,
- |Sn−ℓ|≤ ε.S_n-ell rightvert leq varepsilon .
If the series is convergent, the number ℓdisplaystyle ell (necessarily unique) is called the sum of the series.
Any series that is not convergent is said to be divergent.
Contents
1 Examples of convergent and divergent series
2 Convergence tests
3 Conditional and absolute convergence
4 Uniform convergence
5 Cauchy convergence criterion
6 See also
7 External links
Examples of convergent and divergent series
- The reciprocals of the positive integers produce a divergent series (harmonic series):
- 11+12+13+14+15+16+⋯→∞.displaystyle 1 over 1+1 over 2+1 over 3+1 over 4+1 over 5+1 over 6+cdots rightarrow infty .
- 11+12+13+14+15+16+⋯→∞.displaystyle 1 over 1+1 over 2+1 over 3+1 over 4+1 over 5+1 over 6+cdots rightarrow infty .
- Alternating the signs of the reciprocals of positive integers produces a convergent series:
- 11−12+13−14+15⋯=ln(2)displaystyle 1 over 1-1 over 2+1 over 3-1 over 4+1 over 5cdots =ln(2)
- 11−12+13−14+15⋯=ln(2)displaystyle 1 over 1-1 over 2+1 over 3-1 over 4+1 over 5cdots =ln(2)
- The reciprocals of prime numbers produce a divergent series (so the set of primes is "large"):
- 12+13+15+17+111+113+⋯→∞.displaystyle 1 over 2+1 over 3+1 over 5+1 over 7+1 over 11+1 over 13+cdots rightarrow infty .
- 12+13+15+17+111+113+⋯→∞.displaystyle 1 over 2+1 over 3+1 over 5+1 over 7+1 over 11+1 over 13+cdots rightarrow infty .
- The reciprocals of triangular numbers produce a convergent series:
- 11+13+16+110+115+121+⋯=2.displaystyle 1 over 1+1 over 3+1 over 6+1 over 10+1 over 15+1 over 21+cdots =2.
- 11+13+16+110+115+121+⋯=2.displaystyle 1 over 1+1 over 3+1 over 6+1 over 10+1 over 15+1 over 21+cdots =2.
- The reciprocals of factorials produce a convergent series (see e):
- 11+11+12+16+124+1120+⋯=e.displaystyle frac 11+frac 11+frac 12+frac 16+frac 124+frac 1120+cdots =e.
- 11+11+12+16+124+1120+⋯=e.displaystyle frac 11+frac 11+frac 12+frac 16+frac 124+frac 1120+cdots =e.
- The reciprocals of square numbers produce a convergent series (the Basel problem):
- 11+14+19+116+125+136+⋯=π26.displaystyle 1 over 1+1 over 4+1 over 9+1 over 16+1 over 25+1 over 36+cdots =pi ^2 over 6.
- 11+14+19+116+125+136+⋯=π26.displaystyle 1 over 1+1 over 4+1 over 9+1 over 16+1 over 25+1 over 36+cdots =pi ^2 over 6.
- The reciprocals of powers of 2 produce a convergent series (so the set of powers of 2 is "small"):
- 11+12+14+18+116+132+⋯=2.displaystyle 1 over 1+1 over 2+1 over 4+1 over 8+1 over 16+1 over 32+cdots =2.
- 11+12+14+18+116+132+⋯=2.displaystyle 1 over 1+1 over 2+1 over 4+1 over 8+1 over 16+1 over 32+cdots =2.
- The reciprocals of powers of any n produce a convergent series:
- 11+1n+1n2+1n3+1n4+1n5+⋯=nn−1.displaystyle 1 over 1+1 over n+1 over n^2+1 over n^3+1 over n^4+1 over n^5+cdots =n over n-1.
- 11+1n+1n2+1n3+1n4+1n5+⋯=nn−1.displaystyle 1 over 1+1 over n+1 over n^2+1 over n^3+1 over n^4+1 over n^5+cdots =n over n-1.
- Alternating the signs of reciprocals of powers of 2 also produces a convergent series:
- 11−12+14−18+116−132+⋯=23.displaystyle 1 over 1-1 over 2+1 over 4-1 over 8+1 over 16-1 over 32+cdots =2 over 3.
- 11−12+14−18+116−132+⋯=23.displaystyle 1 over 1-1 over 2+1 over 4-1 over 8+1 over 16-1 over 32+cdots =2 over 3.
- Alternating the signs of reciprocals of powers of any n produces a convergent series:
- 11−1n+1n2−1n3+1n4−1n5+⋯=nn+1.displaystyle 1 over 1-1 over n+1 over n^2-1 over n^3+1 over n^4-1 over n^5+cdots =n over n+1.
- 11−1n+1n2−1n3+1n4−1n5+⋯=nn+1.displaystyle 1 over 1-1 over n+1 over n^2-1 over n^3+1 over n^4-1 over n^5+cdots =n over n+1.
- The reciprocals of Fibonacci numbers produce a convergent series (see ψ):
- 11+11+12+13+15+18+⋯=ψ.displaystyle frac 11+frac 11+frac 12+frac 13+frac 15+frac 18+cdots =psi .
- 11+11+12+13+15+18+⋯=ψ.displaystyle frac 11+frac 11+frac 12+frac 13+frac 15+frac 18+cdots =psi .
Convergence tests
There are a number of methods of determining whether a series converges or diverges.
If the blue series, Σbndisplaystyle Sigma b_n
Comparison test. The terms of the sequence andisplaystyle lefta_nright are compared to those of another sequence bndisplaystyle leftb_nright
. If,
for all n, 0≤ an≤ bndisplaystyle 0leq a_nleq b_n, and ∑n=1∞bndisplaystyle sum _n=1^infty b_n
converges, then so does ∑n=1∞an.displaystyle sum _n=1^infty a_n.
However, if,
for all n, 0≤ bn≤ andisplaystyle 0leq b_nleq a_n, and ∑n=1∞bndisplaystyle sum _n=1^infty b_n
diverges, then so does ∑n=1∞an.displaystyle sum _n=1^infty a_n.
Ratio test. Assume that for all n, an>0displaystyle a_n>0. Suppose that there exists rdisplaystyle r
such that
- limn→∞|an+1an|=r.displaystyle lim _nto infty left
If r < 1, then the series converges. If r > 1, then the series diverges. If r = 1, the ratio test is inconclusive, and the series may converge or diverge.
Root test or nth root test. Suppose that the terms of the sequence in question are non-negative. Define r as follows:
- r=lim supn→∞|an|n,displaystyle r=limsup _nrightarrow infty sqrt[n],
- where "lim sup" denotes the limit superior (possibly ∞; if the limit exists it is the same value).
If r < 1, then the series converges. If r > 1, then the series diverges. If r = 1, the root test is inconclusive, and the series may converge or diverge.
The ratio test and the root test are both based on comparison with a geometric series, and as such they work in similar situations. In fact, if the ratio test works (meaning that the limit exists and is not equal to 1) then so does the root test; the converse, however, is not true. The root test is therefore more generally applicable, but as a practical matter the limit is often difficult to compute for commonly seen types of series.
Integral test. The series can be compared to an integral to establish convergence or divergence. Let f(n)=andisplaystyle f(n)=a_n be a positive and monotonically decreasing function. If
- ∫1∞f(x)dx=limt→∞∫1tf(x)dx<∞,displaystyle int _1^infty f(x),dx=lim _tto infty int _1^tf(x),dx<infty ,
then the series converges. But if the integral diverges, then the series does so as well.
Limit comparison test. If an,bn>0displaystyle lefta_nright,leftb_nright>0, and the limit limn→∞anbndisplaystyle lim _nto infty frac a_nb_n
exists and is not zero, then ∑n=1∞andisplaystyle sum _n=1^infty a_n
converges if and only if ∑n=1∞bndisplaystyle sum _n=1^infty b_n
converges.
Alternating series test. Also known as the Leibniz criterion, the alternating series test states that for an alternating series of the form ∑n=1∞an(−1)ndisplaystyle sum _n=1^infty a_n(-1)^n, if andisplaystyle lefta_nright
is monotonically decreasing, and has a limit of 0 at infinity, then the series converges.
Cauchy condensation test. If andisplaystyle lefta_nright is a positive monotone decreasing sequence, then
∑n=1∞andisplaystyle sum _n=1^infty a_n converges if and only if ∑k=1∞2ka2kdisplaystyle sum _k=1^infty 2^ka_2^k
converges.
Dirichlet's test
Abel's test
Raabe's test
Conditional and absolute convergence
Illustration of the absolute convergence of the power series of Exp[z] around 0 evaluated at z = Exp[i⁄3]. The length of the line is finite.
Illustration of the conditional convergence of the power series of log(z+1) around 0 evaluated at z = exp((π−1⁄3)i). The length of the line is infinite.
For any sequence a1, a2, a3,…displaystyle lefta_1, a_2, a_3,dots right, an≤ |an|displaystyle a_nleq left
for all n. Therefore,
- ∑n=1∞an≤ ∑n=1∞|an|.displaystyle sum _n=1^infty a_nleq sum _n=1^infty left
This means that if ∑n=1∞|an|displaystyle sum _n=1^infty left converges, then ∑n=1∞andisplaystyle sum _n=1^infty a_n
also converges (but not vice versa).
If the series ∑n=1∞|an|displaystyle sum _n=1^infty left converges, then the series ∑n=1∞andisplaystyle sum _n=1^infty a_n
is absolutely convergent. An absolutely convergent sequence is one in which the length of the line created by joining together all of the increments to the partial sum is finitely long. The power series of the exponential function is absolutely convergent everywhere.
If the series ∑n=1∞andisplaystyle sum _n=1^infty a_n converges but the series ∑n=1∞|an|displaystyle sum _n=1^infty left
diverges, then the series ∑n=1∞andisplaystyle sum _n=1^infty a_n
is conditionally convergent. The path formed by connecting the partial sums of a conditionally convergent series is infinitely long. The power series of the logarithm is conditionally convergent.
The Riemann series theorem states that if a series converges conditionally, it is possible to rearrange the terms of the series in such a way that the series converges to any value, or even diverges.
Uniform convergence
Let f1, f2, f3,…displaystyle leftf_1, f_2, f_3,dots right be a sequence of functions.
The series ∑n=1∞fndisplaystyle sum _n=1^infty f_n is said to converge uniformly to f
if the sequence sndisplaystyle s_n of partial sums defined by
- sn(x)=∑k=1nfk(x)displaystyle s_n(x)=sum _k=1^nf_k(x)
converges uniformly to f.
There is an analogue of the comparison test for infinite series of functions called the Weierstrass M-test.
Cauchy convergence criterion
The Cauchy convergence criterion states that a series
- ∑n=1∞andisplaystyle sum _n=1^infty a_n
converges if and only if the sequence of partial sums is a Cauchy sequence.
This means that for every ε>0,displaystyle varepsilon >0, there is a positive integer Ndisplaystyle N
such that for all n≥m≥Ndisplaystyle ngeq mgeq N
we have
- |∑k=mnak|<ε,<varepsilon ,
which is equivalent to
- limn→∞m→∞∑k=nn+mak=0.displaystyle lim _nto infty atop mto infty sum _k=n^n+ma_k=0.
See also
- Convergent sequence
- Normal convergence
- List of mathematical series
- Divergent series
External links
Hazewinkel, Michiel, ed. (2001) [1994], "Series", Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4.mw-parser-output cite.citationfont-style:inherit.mw-parser-output qquotes:"""""""'""'".mw-parser-output code.cs1-codecolor:inherit;background:inherit;border:inherit;padding:inherit.mw-parser-output .cs1-lock-free abackground:url("//upload.wikimedia.org/wikipedia/commons/thumb/6/65/Lock-green.svg/9px-Lock-green.svg.png")no-repeat;background-position:right .1em center.mw-parser-output .cs1-lock-limited a,.mw-parser-output .cs1-lock-registration abackground:url("//upload.wikimedia.org/wikipedia/commons/thumb/d/d6/Lock-gray-alt-2.svg/9px-Lock-gray-alt-2.svg.png")no-repeat;background-position:right .1em center.mw-parser-output .cs1-lock-subscription abackground:url("//upload.wikimedia.org/wikipedia/commons/thumb/a/aa/Lock-red-alt-2.svg/9px-Lock-red-alt-2.svg.png")no-repeat;background-position:right .1em center.mw-parser-output .cs1-subscription,.mw-parser-output .cs1-registrationcolor:#555.mw-parser-output .cs1-subscription span,.mw-parser-output .cs1-registration spanborder-bottom:1px dotted;cursor:help.mw-parser-output .cs1-hidden-errordisplay:none;font-size:100%.mw-parser-output .cs1-visible-errorfont-size:100%.mw-parser-output .cs1-subscription,.mw-parser-output .cs1-registration,.mw-parser-output .cs1-formatfont-size:95%.mw-parser-output .cs1-kern-left,.mw-parser-output .cs1-kern-wl-leftpadding-left:0.2em.mw-parser-output .cs1-kern-right,.mw-parser-output .cs1-kern-wl-rightpadding-right:0.2em- Weisstein, Eric (2005). Riemann Series Theorem. Retrieved May 16, 2005.