Calculus II - Series

31 flashcards · Mathematics · by Harrison Holbrook

Infinite series
If a1,a2,...an,...a_1,a_2,...a_n,...a1,a2,...an,... is an infinite collection of numbers, the expression ∑k=1∞ak=a1+a2...\sum_{k=1}^\infty a_k=a_1+a_2...∑k=1∞ak=a1+a2... is called an infinite series, or simply a series
Partial sums
The partial sums (Capital SnS_nSn ) of a series ∑k=1∞ak\sum_{k=1}^\infty a_k∑k=1∞ak defined by the sequence sns_nsn is Sn=∑k=1nskS_n=\sum_{k=1}^n s_kSn=∑k=1nsk , where nnn is the number of terms of the partial sum
Convergence of a series from its partial sums
If the sequence SnS_nSn of partial sums of an infinite series ∑k=1∞ak\sum_{k=1}^\infty a_k∑k=1∞ak has a limit SSS , then the series converges and is said to have a sum SSS . That is, if lim⁡n→∞Sn=S\lim_{n\to\infty} S_n=Slimn→∞Sn=S , then ∑k=1∞ak=S\sum_{k=1}^\infty a_k=S∑k=1∞ak=S If lim⁡n→∞Sn\lim_{n\to\infty} S_nlimn→∞Sn is infinite or does not exist, the series diverges
Telescoping series
A series where the middle terms of the series expansion cancel
∑k=1nk\sum_{k=1}^n k∑k=1nk
∑k=1nk=n(n+1)2\sum_{k=1}^n k=\frac{n(n+1)}{2}∑k=1nk=2n(n+1)
Geometric series
A series of the form ∑k=0∞ark=∑k=1∞ark−1=a+ar+ar2+ar3...\sum_{k=0}^\infty ar^k = \sum_{k=1}^\infty ar^{k-1}=a+ar+ar^2+ar^3...∑k=0∞ark=∑k=1∞ark−1=a+ar+ar2+ar3... where a≠0a \neq 0a=0 is called a geometric series with ratio rrr
Convergence of a geometric series
If ∣r∣<1|r|<1∣r∣<1 , the geometric series ∑k=1∞ark−1\sum_{k=1}^\infty ar^{k-1}∑k=1∞ark−1 converges, and its sum is ∑k=1∞ark−1=a1−r\sum_{k=1}^\infty ar^{k-1}=\frac{a}{1-r}∑k=1∞ark−1=1−ra . If ∣r∣≥1|r| \geq 1∣r∣≥1 , the geometric series ∑k=1∞ark−1\sum_{k=1}^\infty ar^{k-1}∑k=1∞ark−1 diverges
Harmonic series
The series ∑k=1∞1k=11+12+13...\sum_{k=1}^\infty \frac{1}{k}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}...∑k=1∞k1=11+21+31... is the Harmonic Series, and is known to diverge. It is a special case of a p-series where p=1p=1p=1 .
Limit of the sequence of a convergent infinite series
If the infinite series ∑k=1∞ak\sum_{k=1}^\infty a_k∑k=1∞ak converges, lim⁡n→∞an=0\lim_{n\to\infty} a_n=0limn→∞an=0
Corollary of a limit of a sequence as nnn approaches infinity
If lim⁡n→∞an=0\lim_{n\to\infty} a_n=0limn→∞an=0 , then the series ∑k=1∞ak\sum_{k=1}^\infty a_k∑k=1∞ak may converge or diverge (tells us nothing).
Limit of sequence test for divergence.
If lim⁡n→∞an≠0\lim_{n\to\infty} a_n \neq 0limn→∞an=0 , the series ∑k=1∞ak\sum_{k=1}^\infty a_k∑k=1∞ak diverges.
Two infinite series identical after a certain turn (same partial sums after this term)
If two infinite series are identical after a certain term, then either both series converge or both series diverge. If both series converge, they do not necessarily have the same sum.
Sum and difference of convergent series.
The sum and difference of two convergent series (or their sequences) is also convergent, and is equal to the sum or difference of their respective series sums
Constant multiple of a convergent series
∑k=1∞cak=c∑k=1∞ak\sum_{k=1}^\infty ca_k=c\sum_{k=1}^\infty a_k∑k=1∞cak=c∑k=1∞ak
General convergence test
An infinite series of positive terms converges if and only if its sequence of partial sums is bounded. The sum of such an infinite series will not exceed an upper bound.
Integral test
Suppose a function fff is continuous, positive, and nonincreasing on the interval [1,∞)[1,\infty)[1,∞) . Also suppose ak=f(k)a_k=f(k)ak=f(k) for all positive integers kkk . If the improper integral ∫1∞f(x)dx\int_1^\infty f(x)dx∫1∞f(x)dx converges, the series ∑k=1∞ak\sum_{k=1}^\infty a_k∑k=1∞ak converges If the improper integral ∫1∞f(x)dx\int_1^\infty f(x)dx∫1∞f(x)dx diverges, the series ∑k=1∞ak\sum_{k=1}^\infty a_k∑k=1∞ak diverges Importantly, the value of the integral is not equal to th…
p-series
A p-series is an infinite series of the form ∑k=1∞1kp=1+12p+13p...\sum_{k=1}^\infty \frac{1}{k^p}=1+\frac{1}{2^p}+\frac{1}{3^p}...∑k=1∞kp1=1+2p1+3p1... where ppp is a positive real number
Convergence and divergence of a p-series
A p-series converges if 1"> p>1p>1p>1 and diverges if 0<p≤10 < p \leq 10<p≤1
Bounds of the sum of a convergent p-series
If 1"> p>1p>1p>1 , then 1p−1<∑k=1∞1kp<1+1p−1\frac{1}{p-1} < \sum_{k=1}^\infty \frac{1}{k^p}< 1 + \frac{1}{p-1}p−11<∑k=1∞kp1<1+p−11
Bounds of the sum of a convergent series
Consider the series ∑k=1∞ak\sum_{k=1}^\infty a_k∑k=1∞ak , where 0"> ak>0a_k>0ak>0 for all positive integers kkk . Suppose fff is a continuous, positive, decreasing function on the interval [1,∞)[1,\infty)[1,∞) and ak=f(k)a_k=f(k)ak=f(k) for all positive integers kkk . If the improper integral ∫1∞f(x)dx\int_1^\infty f(x)dx∫1∞f(x)dx , the sum of the series is bounded by ∫1∞f(x)dx<∑k=1∞ak<a1+∫1∞f(x)dx\int_1^\infty f(x)dx < \sum_{k=1}^\infty a_k < a_1+\int_1^\infty f(x)dx ∫1∞f(x)dx<∑k=1∞ak<…
Comparison test for convergence
If 0<ak≤bk0 < a_k \leq b_k0<ak≤bk for all kkk and ∑k=1∞bk\sum_{k=1}^\infty b_k∑k=1∞bk converges, ∑k=1∞ak\sum_{k=1}^\infty a_k∑k=1∞ak converges by comparison with bkb_kbk
Comparison test for divergence
If 0<ck≤ak0 < c_k \leq a_k0<ck≤ak for all kkk and ∑k=1∞ck\sum_{k=1}^\infty c_k∑k=1∞ck diverges, then ∑k=1∞ak\sum_{k=1}^\infty a_k∑k=1∞ak diverges by comparison with ckc_kck
Limit comparison test
Suppose ∑k=1∞ak\sum_{k=1}^\infty a_k∑k=1∞ak and ∑k=1∞bk\sum_{k=1}^\infty b_k∑k=1∞bk are both series of positive term If lim⁡n→∞anbn=L\lim_{n\to\infty} \frac{a_n}{b_n}=Llimn→∞bnan=L , 0<L<∞0 < L < \infty0<L<∞ , then both series converge or both series diverge if lim⁡n→∞anbn=0\lim_{n\to\infty} \frac{a_n}{b_n}=0limn→∞bnan=0 and if ∑k=1∞bk\sum_{k=1}^\infty b_k∑k=1∞bk converges, then ∑k=1∞ak\sum_{k=1}^\infty a_k∑k=1∞ak converges if lim⁡n→∞anbn=∞\lim_{n\to\infty} \frac{a_n}{b_n} = \in…
Alternating series
An alternating series is either in the form ∑k=1∞(−1)k+1ak=a1−a2+a3−a4...\sum_{k=1}^\infty (-1)^{k+1}a_k=a_1-a_2+a_3-a_4...∑k=1∞(−1)k+1ak=a1−a2+a3−a4... or the form ∑k=1∞(−1)kak=−a1+a2−a3+a4...\sum_{k=1}^\infty (-1)^{k}a_k=-a_1+a_2-a_3+a_4...∑k=1∞(−1)kak=−a1+a2−a3+a4... where 0"> ak>0a_k > 0ak>0 for all integers k≥1k \geq 1k≥1
Alternating series test (a.k.a. Leibniz test, Leibniz Criterion)
If the numbers aka_kak , where 0"> ak>0a_k > 0ak>0 , of an alternating series ∑k=1∞(−1)k+1ak\sum_{k=1}^\infty (-1)^{k+1}a_k∑k=1∞(−1)k+1ak or ∑k=1∞(−1)kak\sum_{k=1}^\infty (-1)^{k}a_k∑k=1∞(−1)kak satisfy the two conditions lim⁡n→∞ak=0\lim_{n\to\infty} a_k=0limn→∞ak=0 aka_kak is nonincreasing ( ak≤ak+1a_k \leq a_{k+1}ak≤ak+1 ) (or eventually nonincreasing) Then the alternating series converges
Ways to verify aka_kak is nonincreasing
an+1−an≤0a_{n+1}-a_n \leq 0an+1−an≤0 for all n≥1n \geq 1n≥1 an+1an\frac{a_{n+1}}{a_n}anan+1 ≤1\leq 1≤1 for all n≥1n \geq 1n≥1 Derivative of a related function for ana_nan is less than or equal to zero for all 0"> x>0x>0x>0
Error bound for a convergent alternating series (that satisfies the two conditions of the alternating series test)
For a convergent series series ∑k=1∞(−1)k+1ak\sum_{k=1}^\infty (-1)^{k+1}a_k∑k=1∞(−1)k+1ak that satisfy the alternating series test, the error EnE_nEn in the sum SnS_nSn of the first nnn terms of the sum SSS is ∣En∣≤an+1|E_n| \leq a_n+1∣En∣≤an+1
Absolute convergence
A series ∑k=1∞ak\sum_{k=1}^\infty a_k∑k=1∞ak is absolutely convergent if the series ∑k=1∞∣ak∣=∣a1∣+∣a2∣...\sum_{k=1}^\infty |a_k| =|a_1|+|a_2|...∑k=1∞∣ak∣=∣a1∣+∣a2∣... is convergent
Absolute convergence test
If a series ∑k=1∞ak\sum_{k=1}^\infty a_k∑k=1∞ak is absolutely convergent, then it is convergent
Conditionally convergent
A series that is convergent but not absolutely convergent is conditionally convergent
Properties of absolutely convergent and conditionally convergent series
If a series is absolutely convergent, any arrangement of its terms results in a series that is also absolutely convergent. Both series will have the sam sums.