2
8/7/2019 Tut2_Soln http://slidepdf.com/reader/full/tut2soln 1/2 Hints/Solutions for Tutorial Problem Set - 2 1. (a) Let x n = 1 n 1+ 1 n and let y n = 1 n for all n N. Then lim n→∞ x n y n = 1 = 0. Since n=1 y n is divergent, by the limit comparison test, n=1 x n is also divergent. (Alternative method: We have 2 n n for all n N. So n 1 n 2 for all n N and hence 1 n 1+ 1 n = 1 n·n 1 n 1 2n for all n N. Since n=1 1 2n is divergent, by the comparison test, the given series is divergent.) (b) We have (log n) log n = (e log(log n) ) log n = (e log n ) log(log n) = n log(log n) for all n 2. Also, log(log n) > 2 for all n > e e 2 . We choose n 0 N such that n 0 > e e 2 . Then 1 (log n) log n = 1 n log(log n) 1 n 2 for all n n 0 . Since n=1 1 n 2 is convergent, by the comparison test, the given series is convergent. (c) For all n N, we have (2n)! n n = 2n n · 2n1 n ··· n+1 n · n! 1. Hence lim n→∞ (2n)! n n = 0. Consequently the given series is divergent. 2. We have 0 < xn 1+xn < x n for all n N. Hence by the comparison test, n=1 xn 1+xn con- verges if  n=1 x n converges. Conversely, let n=1 x n 1+x n converge. Then x n 1+x n 0 and so there exists n 0 N such that xn 1+xn < 1 2 for all n n 0 . This implies that x n < 1 for all n n 0 , i.e. 1 + x n < 2 for all n n 0 and so x n < 2x n 1+x n for all n n 0 . By the comparison test, we conclude that n=1 x n converges. 3. We choose n 0 N such that |α| n 0 < π 2 . Then for all n n 0 , sin( α n ) has the same sign as that of α. Since the sine function is increasing in (0, π 2 ), it follows that the sequence (sin( |α| n )) n=n 0 is decreasing. Also, sin( |α| n ) 0. Hence by Leibniz’s test, n=n0 (1) n sin( α n ) is convergent. Consequently n=1 (1) n sin( α n ) is convergent. Again, n=1 |(1) n sin( α n )| = n=1 | sin( α n )| is divergent by the limit comparison test, since (using lim x0 sin x x = 1) lim n→∞ | sin(α/n)| 1/n = |α| lim n→∞ sin(α/n) α/n = |α| = 0 and n=1 1 n is divergent. Therefore the given series is conditionally convergent. 4. We have a n+1 b n+1 a n bn for all n n 0 . Hence a n bn for all n n 0 , where = a n 0 bn 0 > 0. This implies that a n Mb n for all n n 0 . Since n=1 b n converges, n=1 Mb n converges. Hence by the comparison test, n=1 a n converges. Let b n = n n n!e n for all n N. Then b n+1 b n = 1 e (1 + 1 n ) n for all n N. The sequence ((1+ 1 n ) n+1 ) is decreasing and converges to e. (The fact that (1+ 1 n ) n+1 < (1 + 1 n1 ) n for all n > 1, can be proved by using G.M. > H.M. for the numbers x 1 = 1, x 2 = x 3 = ··· = x n+1 = 1 + 1 n1 .) So (1 + 1 n ) n+1 e for all n N. Hence b n+1 bn n n+1 for all n N. Let a n = 1 n for all n N. Then a n+1 an b n+1 bn for all n N

Tut2_Soln

Embed Size (px)

Citation preview

Page 1: Tut2_Soln

8/7/2019 Tut2_Soln

http://slidepdf.com/reader/full/tut2soln 1/2

Hints/Solutions for Tutorial Problem Set - 2

1. (a) Let xn = 1

n1+1n

and let yn = 1n

for all n ∈ N. Then limn→∞

xnyn

= 1 = 0. Since∞

n=1

yn

is divergent, by the limit comparison test,∞

n=1

xn is also divergent.

(Alternative method: We have 2n ≥ n for all n ∈ N. So n1n ≤ 2 for all n ∈ N and

hence 1

n1+1n

= 1

n·n1n

≥ 12n

for all n ∈ N. Since∞

n=1

12n

is divergent, by the comparison

test, the given series is divergent.)

(b) We have (log n)logn = (elog(log n))logn = (elogn)log(logn) = nlog(logn) for all n ≥ 2.Also, log(log n) > 2 for all n > ee

2. We choose n0 ∈ N such that n0 > ee

2. Then

1(logn)logn

= 1nlog(logn)

≤ 1n2

for all n ≥ n0. Since∞

n=1

1n2

is convergent, by the comparison

test, the given series is convergent.

(c) For all n ∈ N, we have (2n)!nn

= 2nn

· 2n−1n

· · · n+1n

· n! ≥ 1. Hence limn→∞

(2n)!nn

= 0.

Consequently the given series is divergent.

2. We have 0 < xn1+xn

< xn for all n ∈ N. Hence by the comparison test,∞

n=1

xn1+xn

con-

verges if ∞

n=1

xn converges.

Conversely, let∞

n=1

xn1+xn

converge. Then xn1+xn

→ 0 and so there exists n0 ∈ N such

that xn1+xn

< 12

for all n ≥ n0. This implies that xn < 1 for all n ≥ n0, i.e. 1 + xn < 2

for all n ≥ n0 and so xn < 2xn1+xn for all n ≥ n0. By the comparison test, we conclude

that∞

n=1

xn converges.

3. We choose n0 ∈ N such that |α|n0

< π2

. Then for all n ≥ n0, sin(αn

) has the same

sign as that of  α. Since the sine function is increasing in (0, π2

), it follows that the

sequence (sin( |α|n

))∞n=n0 is decreasing. Also, sin(|α|n

) → 0. Hence by Leibniz’s test,∞

n=n0

(−1)n sin(αn

) is convergent. Consequently∞

n=1

(−1)n sin(αn

) is convergent.

Again,∞

n=1

|(−1)n sin(αn

)| =∞

n=1

| sin(αn

)| is divergent by the limit comparison test, since

(using limx→0

sinxx

= 1) limn→∞

| sin(α/n)|1/n

= |α| limn→∞

sin(α/n)α/n

= |α| = 0 and

n=1

1n

is divergent.

Therefore the given series is conditionally convergent.

4. We have an+1

bn+1≤ an

bnfor all n ≥ n0. Hence an

bn≤ M  for all n ≥ n0, where M  =

an0bn0

> 0.

This implies that an ≤ M bn for all n ≥ n0. Since∞

n=1

bn converges,∞

n=1

M bn converges.

Hence by the comparison test,∞

n=1

an converges.

Let bn = nn

n!enfor all n ∈ N. Then bn+1

bn= 1

e(1 + 1

n)n for all n ∈ N. The sequence

((1+ 1n)n+1) is decreasing and converges to e. (The fact that (1 + 1

n)n+1 < (1 + 1n−1)n

for all n > 1, can be proved by using G.M. > H.M. for the numbers x1 = 1,x2 = x3 = · · · = xn+1 = 1 + 1

n−1.) So (1 + 1

n)n+1 ≥ e for all n ∈ N. Hence

bn+1

bn≥ n

n+1for all n ∈ N. Let an = 1

nfor all n ∈ N. Then an+1

an≤ bn+1

bnfor all n ∈ N

Page 2: Tut2_Soln

8/7/2019 Tut2_Soln

http://slidepdf.com/reader/full/tut2soln 2/2

and∞

n=1

an is divergent. Therefore by the above result,∞

n=1

bn must also be divergent.

5. Let an = (−1)n(x−1)n

2nn2for all n ∈ N. Then lim

n→∞|an+1

an| = 1

2|x − 1|. Hence by the ratio

test,∞

n=1

an converges (absolutely) if  12

|x − 1| < 1, i.e. if  x ∈ (−1, 3) and diverges if 

1

2

|x − 1| > 1, i.e. if  x ∈ (−∞, −1) ∪ (3, ∞). If  1

2

|x − 1| = 1, i.e. if  x ∈ {−1, 3}, then∞

n=1

|an| =∞

n=1

1n2

converges and hence∞

n=1

an converges. Therefore the set of  x ∈ R for

which∞

n=1

an converges is [−1, 3].