Sequences and series: exercise sheet 5

Exercise.

For each of the following, decide (with justification) which of the following series converge.

(a) n = 1 n + 1 n 3 - 2

(b) n = 1 n - 1 n 2 - 2

(c) n = 1 n 2 + n - 1 n 3

(d) n = 1 n 2 + n - 1 n 4 - 3 n 2 + 1

(e) n = 1 ( n + 1 ) 2 - n 2 n 3 - n 2 + 17

(f) n = 2 ( n 2 + n - 1 ) 1 2 ( n 5 - 2 ) 1 2

(g) n = 2 ( n 2 + n - 1 ) 1 2 ( n 3 - 2 ) 1 2

(h) n = 1 n 2 n (Use the ratio test.)

(i) n = 1 n n n (You may find it helpful to recall that ( 1 + 1 n ) n e = 2.71828... as n .)

(j) n = 1 2 n + 3 n 4 n + 5 n

Exercise.

(a) Prove that n = 1 n x n diverges for all x > 0 . (Use the null sequence test.)

(b) Prove that n = 1 n n 2 + 1 x n diverges for all x > 1 and converges for 0 < x < 1 . (Use the ratio test.)

(c) Prove that n = 1 x n n converges for all x > 0 .

Exercise.

Let a , b > 0 . Show that

(a) n = 1 a + n b + n x n converges when 0 < x < 1 and diverges when x 1 .

(b) a b x + 2 ( a + 1 ) b 2 x 2 + 3 ( a + 2 ) b 3 x 3 + converges when 0 < x < b and diverges when x b .

Exercise.

Discuss the convergence of

1 2 x + 1 · 3 2 · 5 x 2 + 1 · 3 · 5 2 · 5 · 8 x 3 + 1 · 3 · 5 · 7 2 · 5 · 8 · 11 x 4 +

Solution to selected parts of these exercises

Exercise 5.1(a)

The series has positive terms for n 2 and

n + 1 n 3 - 2 2 n n 3 / 2 = 4 n 2

for n 2 , so n = 2 n + 1 n 3 - 2 converges by comparison test with n = 2 1 n 2 hence n = 1 n + 1 n 3 - 2 also converges. (OR: you could use the limit comparison test here.)

Exercise 5.1(b)

The series has positive terms for n 2 and

n - 1 n 2 - 2 n / 2 2 n 2 = 1 4 1 n

for n 2 , so n = 2 n - 1 n 2 - 2 diverges by comparison test with n = 2 1 n hence n = 1 n - 1 n 2 - 2 also diverges. (OR: you could use the limit comparison test here.)

Exercise 5.3(a)

For 1 > x > 0 , the terms in the series n = 1 a + n b + n x n are all positive. Let a n = a + n b + n x n . We apply the ratio test.

a n + 1 a n = a + n + 1 b + n + 1 x n + 1 b + n a + n x - n = ( a n + 1 + 1 n ) ( b n + 1 + 1 n ) ( b n + 1 ) ( a n + 1 ) x

which converges to x by continuity of the arithmetic operations and 1 n 0 . Therefore by the ratio test, the series converges if x < 1 and diverges if x > 1 . If x = 1 the series is n = 1 a + n b + n and a + n b + n = a / n + 1 b / n + 1 1 so the series diverges by the null sequence test.