Tut2_Soln

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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

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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].