MATH 2300-004 QUIZ 6 Name:

1. What is a sequence? What does it mean for a sequence to converge? Give an example of a convergent sequence and of a divergent sequence.

A sequence (an)n is an ordered list of real numbers. A sequence converges,

lim an = L, n→∞

if there is a real number L to which the an are approaching as n increases, i.e. an becomes arbitrarily close to L for n sufficiently large.

[More precisely, the sequence an converges to L if for any error  > 0 (no matter how small!), there is an index N > 0 (possibly very big!) such that |an − L| <  when n ≥ N. In other words, when you go out further than N terms in the sequence, the terms of the sequence are within the specificed small error around L.] Here are some convergent sequences and their limits:  ∞   ∞ 1 1 1 1 1 (an)n=1 = = 1, , , , ,... , n n=1 2 3 4 5 1 lim an = lim = 0 n→∞ n→∞ n  √ q p √  ∞ p ∞   π  1 2 + 2 2 + 2 + 2 (bn)n=0 = cos = −1, 0, √ , , ,... , 2n n=0 2 2 2  π   π  lim bn = lim cos = cos lim = cos(0) = 1, n→∞ n→∞ 2n n→∞ 2n  n ∞   ∞ (−1) 3 2 5 4 (cn)n=1 = 1 + = 0, , , , ,... , n n=1 2 3 4 5 (−1)n lim cn = lim 1 + = 1. n→∞ n→∞ n

Here are some divergent sequences:

∞ n ∞ (an)n=0 = (2 )n=0 = (1, 2, 4, 8, 16,...) exponential growth, √ √ √ √ ! ∞ ∞  nπ  3 3 3 3 (bn)n=0 = sin = 0, , , 0, − , − ,... 3 n=0 2 2 2 2 periodic sequence, ∞ n 2
∞ (cn)n=0 = (−1) (n + 1) n=0 = (1, −2, 5, −10, 17,...) grows in absolute value, alternating in sign. 2. What is a series? What does it mean for a series to converge? Give an example of a convergent series and of a divergent series.

∞ X A series an is (a formal expression for) an attempt to add infinitely many numbers n=0 an, the terms of the series. A series is convergent if the limit of the partial sums

N X lim an = lim (a0 + a1 + ... + aN−1 + aN ) N→∞ N→∞ n=1 exists. In other words, we add up finitely many terms (in order) from a given sequence an and see if the partial sums approach a limiting value. Here are some convergent series:

∞ X 1 1 1 1 = 1 + + + ... = = 2, 2n 2 4 1 − 1/2 n=0 a convergent geometric series, r = 1/2, |r| < 1.

∞ X 1 1 1 1 + + + ... = 1, n(n + 1) 2 6 12 n=1

X 1 a telescoping series, or convergent by comparison to . n2 n

∞ X 1 1 1 π2 = 1 + + + ... = , n2 4 9 6 n=1

a convergent p-series, p = 2 > 1. [The fact that the sum is π2/6 is a non-trivial fact we won’t show in class.] Here are some divergent series:

∞ X 2n, a divergent geometric series, r = 2, |r| ≥ 1, n=0 ∞ X 1 √ , a divergent p-series, p = 1/2 ≤ 1, n n=1 ∞ X 2n2 − 1 X 1 , divergent by comparison to the harmonic series . 3n3 + 1 n n=0 n 3. Use the integral test to determine the convergence or divergence of the series ∞ X 1 . n(ln n)2 n=2

1 The function f(x) = x(ln x)2 is positive and decreasing on the interval [2, ∞) (positive because x and (ln x)2 are positive there, decreasing because the numerator is fixed and the denominator is increasing). The series therefore behaves like the improper integral

Z ∞ dx Z T dx Z ln T du 2 = lim 2 = lim 2 2 x(ln x) T →∞ 2 x(ln x) T →∞ ln 2 u 1 ln T 1 1 1 = lim − = lim − + = < ∞. T →∞ u ln 2 T →∞ ln T ln 2 ln 2 The improper integral converges, therefore the series converges as well.

4. Use the comparison test to determine the convergence or divergence of the series ∞ X 3n − n3 . 5n + n5 n=0

We can use either the direct comparison test or the limit comparison test, in both cases ∞ X comparing to the convergent geometric series (3/5)n. n=0 For the direct comparison test, we have

3n − n3 3n 0 ≤ ≤ , 5n + n5 5

so the series in the problem statement is dominated by the convergent geometric series P n n(3/5) . For the limit comparison test, we have

3n−n3 n3 n 5 1 − lim 5 +n = lim 3n = 1 n→∞ (3/5)n n→∞ n5 1 + 5n since n5 n3 lim = lim = 0 n→∞ 5n n→∞ 3n e.g. using L’Hˆopital’srule a few times. Therefore the series in the problem statement P n has the same convergence behavior as the convergent geometric series n(3/5) .