Mathematics 242 – Principles of Analysis Selected Solutions – Problem Set 5 October 26, 2004

Section 17/4a: First we show that if lim_{n→∞} s_{n} = s and k is a constant, then lim_{n→∞} ks_{n} = ks. Let ε > 0. Then since lim_{n→∞} s_{n} = s, there exists an N such that |s_{n} − s| < ε/|k| for all n ≥ N. But then for all n > N (the same N),

|ks_{n} − ks| = |k||s_{n} − s|
< ε/|k| · |k|
= ε

This shows that lim_{n→∞} ks_{n} = ks.

Next, we show that then lim_{n→∞} k + s_{n} = k + s. As above, given any ε > 0, there exists an N such that |s_{n} − s| < ε for all n ≥ N. But then for all n > N,

\begin{aligned}|(k + s_{n}) − (k + s)| & = |s_{n} − s| \\ & < ε \end{aligned}

This shows that lim_{n→∞} k + s_{n} = k + s.

Section 17/13: We assume that lim_{n→∞} s_{n} = s and lim_{n→∞} t_{n} = t. Then for all ε > 0, there exist N_{1} such that |s_{n} − s| < \sqrt{ε/3} for all n ≥ N_{1}. Similarly, there exists N_{2} such that |t_{n} − t| < \sqrt{ε/3} for all n ≥ N_{2}. Moreover, there exist N_{3} such that |s_{n} − s| < ε/(3|t|) for all n ≥ N_{3} and N_{4} such that |t_{n} − t| < ε/(3|s|) for all n ≥ N_{4}. Then for all n > max(N_{1}, N_{2}, N_{3}, N_{4}), we have (using the triangle inequality on the identity given in the problem)

|s_{n}t_{n} − st| ≤ |s_{n} − s||t_{n} − t| + |s||t_{n} − t| + |t||s_{n} − s|
< (\sqrt{ε/3})^{2} + |s|ε/(3|s|) + |t|ε/(3|t|)
= ε/3 + ε/3 + ε/3
= ε

This shows that lim_{n→∞} s_{n}t_{n} = st.

Section 18/3d: Let s_{n} be the sequence defined by s_{1} = 2, s_{n+1} = \sqrt{2s_{n} + 1} for all n ≥ 1. First we show that (s_{n}) is bounded above by 3 by mathematical induction on n. The base case is n = 1, and s_{1} = 2 < 3 is given. Next, assume s_{k} < 3. Then s_{k+1} = \sqrt{2s_{n} + 1} < \sqrt{6 + 1} = \sqrt{7} < 3. Hence s_{n} < 3 for all n ≥ 1.

Next, we show that s_{n} is monotone increasing (i.e. s_{n} ≤ s_{n+1} for all n ≥ 1), by induction again. The base case is s_{1} = 2 ≤ s_{2} = \sqrt{5} which is clearly true. Assume s_{k} ≤ s_{k+1}. By the order properties of the real numbers, this implies 2s_{k} + 1 ≤ 2s_{k+1} + 1, which implies s_{k+1} = \sqrt{2s_{k} + 1} ≤ \sqrt{2s_{k+1} + 1} = s_{k+2}. Hence s_{n} ≤ s_{n+1} for all n.

Finally, we want to identify the limit s. Taking lim_{n→∞} in the equation: s_{n+1} = \sqrt{2s_{n} + 1} we get s = \sqrt{2s + 1}, which shows that s must be a solution of the quadratic polynomial equation s^{2} − 2s − 1 = 0. The quadratic formula gives s = 1 ± \sqrt{2}. The negative sign gives a negative value, which is not possible since s_{n} ≥ 0 for all n. Hence s = 1 + \sqrt{2} ≐ 2.414.

Section 18/10: We let s_{n} = (1 + \frac{1}{n})^{n}. First, we show that s_{n} is bounded above by 3. By the binomial theorem we have

(1 + \frac{1}{n})^{n} = ∑_{k=0}^{n} \binom{n}{k}\frac{1}{n^{k}}

= 1 + 1 + \frac{1}{2!} \frac{n(n − 1)}{n^{2}} + \frac{1}{3!} \frac{n(n − 1)(n − 2)}{n^{3}} + ⋯

≤ 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + ⋯ + \frac{1}{n!}

≤ 1 + 1 + \frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + ⋯ + \frac{1}{2^{n−1}}

< 3

(Explanation: from the first line to the second, we use the definition of the binomial coefficient \binom{n}{k}, then note that \frac{n(n−1)···(n−k+1)}{n^{k}} ≤ 1 for all k to get the next line. Replacing all factors > 1 in the factorials by 2 gives a larger result. Finally we use the result from 18.6 to sum the geometric series with ratio 1/2 (there’s an extra term of 1 at the start).

Next, we need to see that s_{n} is a monotone increasing sequence. This can be done by comparing the second line above:

1+1+\frac{1}{2!} \frac{n(n − 1)}{n^{2}}+\frac{1}{3!} \frac{n(n − 1)(n − 2)}{n^{3}}+⋯ = 1+1+\frac{1}{2!} (1 − \frac{1}{n})+\frac{1}{3!} (1 − \frac{1}{n}) (1 − \frac{2}{n})+⋯

to what we get in a similar fashion from s_{n+1}:

\begin{matrix}1 + 1 + \frac{1}{2!} \frac{(n + 1)n}{(n + 1)^{2}} + \frac{1}{3!} \frac{(n + 1)n(n − 1)}{(n + 1)^{3}} + ⋯ \\ = 1 + 1 + \frac{1}{2!} (1 − \frac{1}{n + 1}) + \frac{1}{3!} (1 − \frac{1}{n + 1}) (1 − \frac{2}{n + 1}) + ⋯ \end{matrix}

(There is one more term in the sum for n + 1 than for n as well.) Note that n + 1 > n, so (1 − k/n) < (1 − k/(n + 1)) for all k. Hence s_{n+1} > s_{n} for all n. Since the sequence s_{n} is monotone increasing and bounded above, it converges. The limit is (by definition) the number e (base of the natural logarithms).