Solutions to sample midterm questions

1. Let x = 1 + \sqrt{1 + \sqrt{2}} . Then

x − 1 = \sqrt{1 + \sqrt{2}}
(x − 1)^{2} = 1 + \sqrt{2}
x^{2} − 2x + 1 = 1 + \sqrt{2}
x^{2} − 2x = \sqrt{2}
x^{4} − 4x^{3} + 4x^{2} = 2
x^{4} − 4x^{3} + 4x^{2} − 2 = 0

Hence if x = p/q, then p divides 2 and q divides 1. The only possibilities are ±1, and ±2. But \sqrt{1 + \sqrt{2}} > 1, and thus x > 2. Thus x must be irrational.

2. Define a_{n} = 8^{n}/(n!)^{2}. Then

\frac{a_{n+1}}{a_{n}} = \frac{8^{n+1}}{((n + 1)!)^{2}} \frac{(n!)^{2}}{8^{n}}
= \frac{8}{(n + 1)^{2}} → 0

and therefore by the ratio test, ∑ 8^{n}/(n!)^{2} converges.

Now consider ∑ (−1)^{n}b_{n} where b_{n} = \sqrt{n^{2} + n} . Since n and n^{2} are both increasing functions, n^{2}+ n is an increasing function also, and hence 1/\sqrt{n^{2} + n} is a decreasing function. In addition, b_{n} < 1/n for all n, so b_{n} → 0 as n → ∞. Therefore, the series ∑ (−1)^{n}b_{n} satisfies the conditions for the alternating series theorem, and hence it converges.

3. (a) Let x ∈ S ∪ T. Then either x ∈ S so x ≤ sup S, or x ∈ T so x ≤ sup T. Hence, x ≤ max{sup S, sup T}. Thus max{sup S, sup T} is an upper bound for sup S ∪ T. Now suppose that m is an upper bound for S ∪ T. Hence m ≥ x for all x ∈ S ∪ T. Thus m ≥ s for all s ∈ S, so m ≥ sup S as sup S is the least upper bound for S. Similarly m ≥ t for all t ∈ T. Hence m ≥ sup T as sup T is the least upper bound of T. Therefore m ≥ max{sup S, sup T}. Hence max{sup S, sup T} is an upper bound, and it is the least upper bound, so it must equal sup S ∪ T. Now consider x ∈ S ∩ T. Hence x ∈ S and x ∈ T. Then x ≤ sup S and x ≤ sup T, so x ≤ min{sup S, sup T}, and therefore sup S ∩ T ≤ min{sup S, sup T}.

(b) For a non-empty set A, sup A ≠ −∞, so it suffices to consider when the suprema become positive infinity. Suppose sup S = ∞. Then S is not bounded above. Hence S ∪ T is not bounded above. Therefore sup S ∪ T = ∞ and the identity still holds. For the second identity, if sup S = ∞, then min{sup S, sup T} = sup T. Since sup T is an upper bound for T, it is also an upper bound for S ∩ T, and hence the identity still holds. The same arguments can be applied if sup T = ∞.

(c) Consider S = {1, 3} and T = {1, 2}. Then sup S = 3 and sup T = 2, so min{sup S, sup T} = 2. However, S ∩ T = {1} and so sup S ∩ T = 1 < 2.

4. Let lim s_{n} = s. Since s_{n} converges, there exists an N_{1} such that n > N_{1} implies that |s_{n} − s| < 1. Hence −1 < s_{n} − s and s_{n} > s − 1. Now pick M > 0. Since t_{n} diverges, there exists an N_{2} such that

t_{n} > 1 − s + M

for all n > N_{2}. Hence for n > max{N_{1}, N_{2}},

s_{n} + t_{n} > (s − 1) + 1 − s + M = M

and thus s_{n} + t_{n} diverges to infinity.

5. (a) Define a_{N} = sup{s_{n} : n > N} and b_{N} = sup{t_{n} : n > N}. Now, for n > N,

s_{n} + t_{n} ≤ a_{N} + b_{N}

since a_{N} and b_{N} are upper bounds for s_{n} and t_{n}. If c_{N} = sup{s_{n} + t_{n} : n > N}, then

c_{N} ≤ a_{N} + b_{N}.

The sequences (a_{N}), (b_{N}), and (c_{N}) are non-increasing. Suppose that lim c_{N} > lim a_{N} + lim b_{N}. Then lim c_{N} = lim a_{N} + lim b_{N} + ϵ for some ϵ > 0, so there exist K_{1} and K_{2} such that if k > K_{1}

a_{k} < lim a_{N} + \frac{ϵ}{3}

and if k > K_{2} then

b_{k} < lim b_{N} + \frac{ϵ}{3}.

Now for all k > max{K_{1}, K_{2}},

c_{k} ≤ a_{k}+ b_{k}
< ( lim a_{N} + \frac{ϵ}{3}) + ( lim b_{N} + \frac{ϵ}{3})
< (lim a_{N} + lim b_{N} + ϵ) − \frac{ϵ}{3} = lim c_{N} − \frac{ϵ}{3}.

But then |c_{k} − lim c_{N}| > \frac{ϵ}{3} for all k > max\{K_{1}, K_{2}\}, so c_{k} does not converge to lim c_{N} which is a contradiction. Hence lim c_{N} ≤ lim a_{N} + lim b_{N}, and hence lim sup s_{n} + t_{n} ≤ lim sup s_{n} + lim sup t_{n}.

(b) Suppose

s_{n} = \{ \begin{matrix}1 & if n is even \\ 0 & if n is odd \end{matrix}

and that

t_{n} = \{ \begin{matrix}0 & if n is even \\ 1 & if n is odd. \end{matrix}

Then sup{s_{n} : n > N} and sup{t_{n} : n > N} = 1 for all N ∈ , and hence

(lim sup s_{n}) · (lim sup t_{n}) = 1 · 1 = 1.

However, s_{n}t_{n} = 0 for all n, and thus lim sup(s_{n}t_{n}) = 0 ≠ 1.

6. Suppose that a > b. Then define ϵ = a − b > 0. Then there exists an N_{1} such that n > N_{1} implies that |a_{n} − a| < ϵ/2. Similarly there exists an N_{2} such that n > N_{2} implies that |b_{n} − b| < ϵ/2. Now consider any k such that k > max{N_{1}, N_{2}}. Then |a_{k}− a| < ϵ/2, and hence −ϵ/2 < a_{k}− a, so

a_{k} > a − \frac{ϵ}{2} = a − \frac{a − b}{2} = \frac{a + b}{2} .

In addition, |b_{k}− b| < ϵ/2, so b_{k}− b < ϵ/2, and hence

b_{k} < b + \frac{ϵ}{2} = b + \frac{a − b}{2} = \frac{a + b}{2} .

Combining these two inequalities shows that a_{k} > b_{k}, which is a contradiction. Thus a ≤ b.