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HW5, q + a - Homework assignment 5

Homework assignment 5

Course

Real Analysis I (MATH 370)

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MATH 370: Homework 5

Due Tuesday, 3/7 in class

Reading: Ross 1, 1.

Determine which of the following series converge. Justifyyour answers.

(a)

∑n 2 2 n (b)

∑ 2 n n! (c)

∑ 1

log(1 +n)

(d)

∑n 2 nn (e)

∑n! nn 2. (Ross, p: Exercise 14) Prove that if

∑

an and

∑

bn are convergent series of nonnegative numbers, then

∑√

anbnconverges. (Hint: Show

√

anbn≤an+bn.)

(Ross, p: Exercise 15) Show that

∑∞

n= 1 n(logn)pconverges if and only ifp >1. 4. Determine which of the following series converge. Justifyyour answers.

(a)

∑∞

2 n+ 3 n 2 +n+ 1

(b)

∑∞

2 n+ 3 n 3 +n+ 1

(c)

∑∞

logn n

(d)

∑∞

logn n 2

(e)

∑∞

1

n(logn)(log logn)

Determine which of the following series converge. Justifyyour answers.

(a)

∑∞

(−1)n logn

(b)

∑∞

(−1)n

√

n− 1 n+ 1

(c)

∑∞

(−1)nn! nn

BONUS/Optional: (Ross, p: Exercise 15) Prove that for a decreasing sequence of real numbers,

∑

anconverges implies limn→∞nan= 0. Does the converse hold?

(c)

∑∞

logn n

≥

∑

log 2 n

=∞; hence, diverges.

(d)

∑∞

logn n 2

≤

∑

M

n 1. 5 <∞whereM >0 exists because the sequence logn n has a

limit (= 0); hence, the original series converges.

(e)

∑∞

1

n(logn)(log logn)

diverges b/c

∫∞

dx x(logx)(log logx)

= log log logx

∣∣

∣

∞ =∞.

Determine which of the following series converge. Justifyyour answers.

(a)

∑∞

(−1)n logn converges by the alternating series test

(b)

∑∞

(−1)n

√

n− 1 n+ 1 converges because the series is alternating and the absolute

value of the terms is eventually decreasing (because the log√ arithmic derivative of n− 1 n+ 1

1

2 n− 2

+

1

n+ 1

>0 forn≥2) to zero, so that the alternating series test applies to an appropriate tail of the original sequence.

(c)

∑∞

(−1)nn! nn converges because n! nn

>

(n+ 1)! (n+ 1)n+ holds for alln, and n! nn

≤

1 →0,

so that the alternating series test applies directly.

BONUS/Optional: (Ross, p: Exercise 15) Prove that for a decreasing sequence of real numbers,

∑

anconverges implies limn→∞nan= 0. Does the converse hold? Observe that eachan is positive since (an) is decreasing and liman = 0. Given any ε >0, convergence of

∑

anallows us to choose anMso thataM+1+aM+2+···< ε/2. Now ifN > Mis chosen so thatan< ε 2 M for alln > N, thenn > Nimplies

nan=Man+ (n−M)an< ε 2

+aN+1+aN+2+···+an< ε.

Hence, limnan= 0. The converse does not hold as seen by consideringan=

1

nlogn

.

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