X n=1 1 n3 is convergent

1. Q3

(1) Test the convergence/divergence of the following infinite series.

(a)

∞

X

n=1

 1 + 1

n

n

. Sol: Since lim

n→∞

 1 + 1

n

n

= e 6= 0, by n-th term test, the series is divergent.

(b)

∞

X

n=10

1 n(ln n)³.

Sol: The sequence a_n= 1/n(ln n)³is decreasing and nonnegative and lim

n→∞

1 n(ln n)³ = 0. By the Cauchy condensation principle, the convergence of the series is equiv- alent to the convergence of the series

∞

X

n=1

2ⁿ· 1 2ⁿ(ln 2ⁿ)³ =

∞

X

n=1

1 n³(ln 2)³.

By p-test,

∞

X

n=1

1

n³ is convergent. Hence the original infinite serie is convergent.

(c)

∞

X

n=1

n2ⁿ(n + 1)!

3ⁿn! .

Sol: Let a_n= n2ⁿ(n + 1)!

3ⁿn! for n ≥ 1. Then a_n> 0 and a_n+1

an

= n + 1 n ·2

3 ·n + 2 n + 1. Hence

n→∞lim a_n+1

an

= 2 3 < 1.

By ratio test, the infinite series is convergent.

(d)

∞

X

n=1

 2n − 1 5n + 4

n

.

Sol: Taking the n-th root of the n-th term, we obtain

n

s

 2n − 1 5n + 4

n

= 2n − 1 5n + 4. Taking limit,

n→∞lim

n

s

 2n − 1 5n + 4

n

= lim

n→∞

2n − 1 5n + 4 = 2

5 < 1.

By root test, the series is convergent.

(e)

∞

X

n=1

sin π 2n².

1

2

Sol Since 0 < π 2n² < π

2 for n ≥ 1, 0 < sin π

2n² < π

2n², n ≥ 1.

By p-test

∞

X

n=1

1

n² is convergent. Hence by comparison test, the series is conver- gent.

(2) The infinite series

∞

X

n=1

(−1)ⁿtan1

n is absolutely convergent? conditionally conver- gent? or divergent?

Sol Since the infinite series is an alternating series and (a) (tan 1

n) is decreasing, (b) lim

n→∞tan1 n = 0.

By the Leibnitz test, the series is convergent. Since lim

n→∞

tan_n¹

1 n

= 1 and

∞

X

n=1

1 n is divergent by p-test, by the limit comparison test,

∞

X

n=1

tan1

n is divergent.

Lemma 1.1. The sequence (tan1

n) is decreasing and lim

n→∞tan1 n = 0.

Proof. For n ≥ 1, tan1

n− tan 1

n + 1 = sin_n¹

cos_n¹ − sin_n+1¹ cos_n+1¹

= 1

cos_n¹cos_n+1¹ (sin 1

ncos 1

n + 1 − cos1

nsin 1 n + 1)

= 1

cos_n¹cos_n+1¹ sin 1 n− 1

n + 1

 . Since n ≥ 1,

0 < 1 n− 1

n + 1 < 1 n < π

2. Hence cos1

ncos 1

n + 1 > 0 and sin 1 n− 1

n + 1



> 0. This implies that for n ≥ 1

tan1

n− tan 1

n + 1 > 0.

For n ≥ 1, 1 n < π

2, and hence 0 < tan1

n = sin_n¹ cos_n¹ <

1 n

cos¹_n.

3

We have seen in midterm that lim

n→∞cos1

n = 1. Hence

n→∞lim

1 n

cos_n¹ = lim

n→∞0 = 0.

By the Sandwich principle, we prove lim

n→∞tan1 n = 0.

 (3) Test the convergence/divergence of the infinite series

∞

X

n=1

 tan1

n − sin1 n

 . Sol: For each n ≥ 1, we write

tan1

n− sin 1

n = tan1 n



1 − cos 1 n



= 2 tan1

nsin² 1 2n. Since n ≥ 1, 0 < 1

n < π

2. By sin θ < θ for 0 < θ < π/2, we know that 0 < sin² 1 2n <

1

4n². By p-test,

∞

X

n=1

1

n² is convergent. By comparison test,

∞

X

n=1

sin² 1

2n is convergent.

By the above lemma, lim

n→∞tan1

n = 0 and (tan(1/n)) is positive and decreasing and

n→∞lim tan1

n = 0, by Abel test or Dirichlet test, the infinite series

∞

X

n=1

2 tan1 nsin² 1

2n is convergent.