Problems

---

Solution:

1. Claim \(x_n \geqslant 2 + 4^{n-1}(a-2)\) Proof: (By induction) Base case: Note that when \(n = 1\), \(x_1 = a = 2 + 1 \cdot(a - 2)\). Inductive step, suppose true for some \(n\), then \begin{align*} && x_{n+1} &= x_n^2 - 2 \\ &&&\geq (2+4^{n-1}(a-2))^2 - 2 \\ &&&= 4 + 4^{2n-2}(a-2)^2 + 4^n(a-2) - 2 \\ &&&= 2 + 4^{n}(a-2) + 4^{2n-2}(a-2)^2 \\ &&&\geq 2 + 4^{n+1-1}(a-2) \end{align*} as required,
2. (\(\Leftarrow\)) Suppose \(a > 2\) then \(x_n \geq 2+4^{n-1}(a-2) \to \infty\) as required. Suppose \(a < -2\) then \(x_2 > 4 -2 = 2\) so the sequence starting from \(x_2\) clearly diverges for the same reason. (\(\Rightarrow\)) suppose \(|x_n| \leq 2\) then \(x_{n+1} = x_n^2 - 2 \leq 2\) so the sequence is bounded and cannot tend to \(\infty\).
3. Suppose \(y_n = \frac{Ax_1x_2 \cdots x_n}{x_{n+1}}\) and notice that \(x_{n+1}^2 - 4 = (x_n^2 -2)^2 - 4 = x_n^4 - 4x_n^2 = x_n^2(x_n^2-4)\). In particular, \(\frac{\sqrt{x_{n+1}^2-4}}{x_{n+1}} = \frac{x_n\sqrt{x_n^2-4}}{x_{n+1}} = \frac{x_n x_{n-1} \cdots x_1 \sqrt{x_1^2-4}}{x_{n+1}}\) Therefore if \(A = \sqrt{a^2-4}\) \(y_{n+1} = \frac{\sqrt{x_{n+1}^2-4}}{x_{n+1}}\). Notice that \begin{align*} && y_n &= \frac{\sqrt{x_{n+1}^2-4}}{x_{n+1}} \\ &&&= \sqrt{1 - \frac{4}{x_{n+1}^2}} \to 1 \end{align*}

---

Solution: Claim \(a_n^2+2a_nb_n - b_n^2 = 1\) for all \(n \geq 1\) Proof: (By induction) Base case: (\(n = 1\)). When \(n = 1\) we have \(a_1^2 + 2a_1 b_1-b_1^2 = 1^2+2\cdot1\cdot2-2^2 = 1\) as required. (Inductive step). Now we assume our result is true for some \(n =k\), ie \(a_k^2+2a_kb_k - b_k^2 = 1\), now consider \(n = k+1\) \begin{align*} && a_{k+1}^2+2a_{k+1}b_{k+1} - b_{k+1}^2 &= (a_k+2b_k)^2+2(a_k+2b_k)(2a_k+5b_k) - (2a_k+5b_k)^2 \\ &&&= a_k^2+4a_kb_k+4b_k^2 +4a_k^2+18a_kb_k+20b_k^2 - 4a_k^2-20a_kb_k-25b_k^2 \\ &&&= (1+4-4)a_k^2+(4+18-20)a_kb_k +(4+20-25)b_k^2 \\ &&&= a_k^2+2a_kb_k -b_k^2 = 1 \end{align*} Therefore since our statement is true for \(n = 1\) and when it is true for \(n=k\) it is true for \(n=k+1\) by the POMI it is true for \(n \geq 1\)

1. Notice that \(b_{n+1} \geq 5 b_n\) and therefore \(b_n \geq 5^{n-1} b_1 = 2\cdot 5^{n-1}\), so \begin{align*} && 1 &= a_n^2+2a_nb_n - b_n^2\\ \Rightarrow && \frac1{b_n^2} &= c_n^2 + 2c_n - 1 \\ \to && 0 &= c_n^2 + 2c_n - 1 \quad \text{ as } n\to \infty \\ \end{align*} This has roots \(c = -1 \pm \sqrt{2}\), and since \(c_n > 0\) it must tend to the positive value, ie \(c_n \to \sqrt{2}-1\)
2. Notice that \(c_n^2 + 2c_n - 1 > 0\) so either \(c_n > \sqrt{2}-1\) or \(c_n < -1-\sqrt{2}\), but again, since \(c_n > 0\) we must have \(c_n > \sqrt{2}-1\). Therefore \(\sqrt{2} < c_n + 1\) and \(1+c_n > \sqrt{2} \Rightarrow \frac{1}{1+c_n} < \frac{\sqrt{2}}2 \Rightarrow \frac{2}{1+c_n} < \sqrt{2}\) \begin{array}{c|c|c} n & a_n & b_n \\ \hline 1 & 1 & 2 \\ 2 & 5 & 12 \\ 3 & 29 & 70 \end{array} Therefore \(c_3 = \frac{29}{70}\) and so \(\frac{2}{1 + \frac{29}{70}} = \frac{140}{99} < \sqrt{2} < \frac{29}{70} + 1 = \frac{99}{70}\)

---

Solution: \begin{align*} \frac{r+1}{r} \left(\frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}}\right) &= \frac{r+1}{r} \l \frac{r!(n-1)!}{(n+r-1)!} - \frac{r!n!}{(n+r)!} \r \\ &= \frac{(r+1)!(n-1)!}{r(n+r-1)!} \l 1 - \frac{n}{n+r} \r \\ &= \frac{(r+1)!(n-1)!}{r(n+r-1)!} \frac{r}{n+r} \\ &= \frac{(r+1)!n!}{(n+r)!} \\ &= \frac{1}{^{n+r}\C_{r+1}} \end{align*} \begin{align*} \sum_{n=1}^{\infty}{\frac{1}{^{n+r}\C_{r+1}}} &= \sum_{n=1}^{\infty} \l \frac{r+1}{r} \left(\frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}}\right) \r \\ &= \frac{r+1}{r} \sum_{n=1}^{\infty} \l \frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}} \r \\ &= \frac{r+1}{r} \lim_{N \to \infty} \sum_{n=1}^{N} \l \frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}} \r \\ &= \frac{r+1}{r} \lim_{N \to \infty} \l \frac{1}{^{1+r-1}\C_{r}} - \frac{1}{^{N+r}\C_{r}}\r \\ &= \frac{r+1}{r} \frac{1}{^{1+r-1}\C_{r}} \tag{since \(\frac{1}{^{N+r}\C_{r}} \to 0\)} \\ &= \frac{r+1}{r} \end{align*} When \(r = 2\), we have: \begin{align*} && \frac{3}{2} &= \sum_{n=1}^{\infty}{\frac{1}{^{n+2}\C_{3}}} \\ && &=\frac{1}{^{1+2}\C_{3}} + \sum_{n=2}^{\infty}{\frac{1}{^{n+2}\C_{3}}} \\ && &= 1 + \sum_{n=2}^{\infty}{\frac{1}{^{n+2}\C_{3}}} \\ \Rightarrow && \sum_{n=2}^{\infty}{\frac{1}{^{n+2}\C_{3}}} &= \frac12 \end{align*} \begin{align*} \frac{1}{^{n+1}\C_{3}} &= \frac{3!}{(n+1)n(n-1)} \\ &= \frac{3!}{n^3-n} \\ &> \frac{3!}{n^3} \end{align*} \begin{align*} \frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}} &= \frac{5!}{(n+1)n(n-1)} - \frac{5!}{(n+2)(n+1)n(n-1)(n-2)} \\ &= \frac{5!}{n^3} \frac{n^2}{n^2-1}\l 1- \frac{1}{n^2-4} \r \\ &= \frac{5!}{n^3} \frac{n^2}{n^2-1}\l \frac{n^2-5}{n^2-4} \r \\ &= \frac{5!}{n^3} \frac{n^2(n^2-5)}{(n^2-1)(n^2-4)} \\ &< \frac{5!}{n^3} \end{align*} Since \(k(k-5) < (k-1)(k-4) \Leftrightarrow 0 < 4\), this only makes sense if \(n \geq 3\) \begin{align*} &&\frac{3!}{n^3} &< \frac{1}{^{n+1}\C_{3}} \tag{if \(n \geq 3\)} \\ \Rightarrow &&\sum_{n=3}^\infty \frac{3!}{n^3} &< \sum_{n=3}^\infty \frac{1}{^{n+1}\C_{3}} \\ \Rightarrow && \frac{6}{1^3} + \frac{6}{2^3} + \sum_{n=3}^\infty \frac{3!}{n^3} &< \frac{6}{1^3} + \frac{6}{2^3} + \sum_{n=3}^\infty \frac{1}{^{n+1}\C_{3}} \\ \Rightarrow && \sum_{n=1}^\infty \frac{3!}{n^3} &< 6 + \frac{3}{4} + \sum_{n=2}^\infty \frac{1}{^{n+2}\C_{2+1}} \\ \Rightarrow && \sum_{n=1}^\infty \frac{3!}{n^3} &< 6 + \frac{3}{4} + \frac{1}{2} = \frac{29}{4} \\ \Rightarrow && \sum_{n=1}^\infty \frac{1}{n^3} &< \frac{29}{24} = \frac{116}{96} \\ \end{align*} \begin{align*} && \frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}} &< \frac{5!}{n^3} \\ \Rightarrow && \sum_{n=3}^\infty \l \frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}} \r &< \sum_{n=3}^\infty \frac{5!}{n^3} \\ \Rightarrow && \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=3}^\infty \frac{20}{^{n+1}\C_3} - \sum_{n=3}^\infty \frac{1}{^{n+2}\C_{5}} &< \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=3}^\infty \frac{5!}{n^3} \\ \Rightarrow && \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=2}^\infty \frac{20}{^{n+2}\C_{2+1}} - \sum_{n=1}^\infty \frac{1}{^{n+4}\C_{4+1}} &< \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=3}^\infty \frac{5!}{n^3} \\ \Rightarrow && \frac{120}{1^3} + \frac{120}{2^3} + \frac{20}{2} - \frac{4+1}{4} &< \sum_{n=1}^\infty \frac{5!}{n^3} \\ \Rightarrow && \frac{115}{96} &< \sum_{n=1}^\infty \frac{1}{n^3} \\ \end{align*}

---

Solution: \begin{align*} && \int_{m-\frac12}^\infty \frac{1}{x^2} \d x &= \lim_{K \to \infty} \left [ -x^{-1} \right]_{m-\frac12}^K \\ &&&= \frac{1}{m-\frac12} - \lim_{K \to \infty }\frac{1}K \\ &&&= \frac{1}{m-\frac12} \end{align*}

Notice that \(\displaystyle \frac{1}{r^2} \approx \int_{r-\frac12}^{r+\frac12} \frac{1}{x^2} \d x\) as the area of the orange boxes and under the blue lines are similar.

1. \(\,\) \begin{align*} E &\approx \int_{1-\frac12}^\infty \frac1{x^2} \d x = \frac{1}{1-\frac12} = 2 \\ E &\approx 1 + \int_{2-\frac12}^\infty \frac1{x^2} \d x= 1 + \frac{1}{2 - \frac12} = \frac53 \\ E &\approx 1 +\frac14 + \int_{3-\frac12}^\infty \frac1{x^2} \d x= \frac54 + \frac{1}{3-\frac12} \\ &= \frac54+\frac{2}{5} = \frac{33}{20} \end{align*}
2. The error is \begin{align*} && \epsilon &= \int_{r-\frac12}^{r+\frac12} \frac 1 {x^2} \, \d x - \frac1{r^2} \\ &&&= \frac{1}{r-\frac12} - \frac{1}{r + \frac12} - \frac1{r^2} \\ &&&= \frac{1}{r^2 - \frac14} - \frac1{r^2} \\ &&&= \frac{\frac14}{r^2(r^2-\frac14)} \\ &&&\approx \frac{1}{4r^4} \end{align*} Therefore \begin{align*} && \sum_{n=1}^\infty \frac1{r^4} &\approx 4 \left ( 1 +\frac14 + \int_{3-\frac12}^\infty \frac1{x^2} \d x-\sum_{r=1}^\infty \frac{1}{r^2} \right) + 1 + \frac{1}{2^4}\\ &&&= 4 \left ( \frac{33}{20}-1.645 \right) + 1 + \frac{1}{2^4} \\ &&&= 4 \left ( 1.65-1.645 \right) + 1 + \frac{1}{2^4} \\ &&&= 1.0825 \approx 1.08 \end{align*}

---

Solution:

1. \(\,\) \begin{align*} && \frac{1}{1+x^r} - \frac{1}{1+x^{r+1}} &= \frac{1+x^{r+1}-1-x^r}{(1+x^r)(1+x^{r+1})} \\ &&&= \frac{x^r(x-1)}{(1+x^r)(1+x^{r+1})} \\ \\ && \sum_{r=1}^N \frac{x^r}{(1+x^r)(1+x^{r+1})} &= \sum_{r=1}^N \frac{1}{x-1} \left ( \frac{1}{1+x^r} - \frac{1}{1+x^{r+1}}\right) \\ &&&= \frac{1}{x-1} \Bigg ( \frac{1}{1+x} + \cdots \\ &&& \qquad \qquad \quad - \frac{1}{1+x^2} + \frac{1}{1+x^2} + \cdots \\ &&& \qquad \qquad \quad - \frac{1}{1+x^3} + \frac{1}{1+x^3} + \cdots \\ &&& \qquad \qquad \quad - \cdots \\ &&& \qquad \qquad \quad - \frac{1}{1+x^{N+1}} \Bigg ) \\ &&&= \frac{1}{x-1} \left (\frac{1}{1+x} - \frac{1}{1+x^{N+1}} \right) \\ \\ && \sum_{r=1}^{\infty} \frac{x^r}{(1+x^r)(1+x^{r+1})} &= \lim_{N\to \infty} \frac{1}{x-1} \left (\frac{1}{1+x} - \frac{1}{1+x^{N+1}} \right) \\ &&&= \frac{1}{x-1} \left ( \frac{1}{1+x} - 1\right) \\ &&&= \frac{1}{x-1} \left ( \frac{-x}{1+x} \right) \\ &&&= \frac{x}{1-x^2} \end{align*}
2. \(\,\) \begin{align*} && \sum_{r=1}^\infty \textrm{sech}(ry)\textrm{sech}((r + 1)y) &= \sum_{r=1}^\infty \frac{4}{(e^{ry}+e^{-ry})(e^{(r+1)y}+e^{-(r+1)y})} \\ &&&=\sum_{r=1}^\infty \frac{4e^{-(2r+1)y}}{(1+e^{-2ry})(1+e^{-2(r+1)y})} \\ x = e^{-2y}: &&&= \frac{4e^{-y}e^{-2y}}{1-e^{-4y}} \\ &&&= \frac{4e^{-y}e^{-2y}}{e^{-2y}(e^{2y}-e^{-2y})} \\ &&&=2e^{-y}\textrm{cosech}(2y) \end{align*} \begin{align*} && \sum_{r=-\infty}^\infty \textrm{sech}(ry) \textrm{sech}((r + 1)y) &= \sum_{r=1}^\infty \textrm{sech}(ry) \textrm{sech}((r + 1)y) + \sum_{r=-\infty}^0 \textrm{sech}(ry) \textrm{sech}((r + 1)y) \\ &&&= 2e^{-y}\textrm{cosech}(2y) + \sum_{r=0}^\infty \textrm{sech}(-ry) \textrm{sech}(-(r-1)y) \\ &&&= 2e^{-y}\textrm{cosech}(2y) + \sum_{r=0}^\infty \textrm{sech}((r-1)y) \textrm{sech}(ry) \\ &&&= 4e^{-y}\textrm{cosech}(2y) + \textrm{sech}(y) + \textrm{sech}(-y) \\ &&&= 4e^{-y}\textrm{cosech}(2y)+2\textrm{sech}(y) \\ &&&= 4e^{-y} \frac12 \textrm{sech}(y) \textrm{cosech}(y) + 2 \textrm{sech}(y) \\ &&&= 2\textrm{sech}(y) \left ( e^{-y} \textrm{cosech}(y)+1 \right) \\ &&&= 2\textrm{sech}(y) \left ( \frac{2}{e^{2y}-1} + 1 \right) \\ &&&= 2\textrm{sech}(y) \left ( \frac{e^{2y}+1}{e^{2y}-1} \right) \\ &&&= 2 \textrm{cosech}(y) \end{align*}

---

Solution: \begin{align*} && \sum_{r=k^n}^{k^{n+1}-1} f(r) &\leq \sum_{r=k^n}^{k^{n+1}-1} f(k^{n}) \\ &&&= (k^{n+1}-k^n)f(k^n) \\ &&&= k^n(k-1)f(k^n) \\ \\ && \sum_{r=k^n}^{k^{n+1}-1} f(r) &\geq \sum_{r=k^n}^{k^{n+1}-1} f(k^{n+1}) \\ &&&= (k^{n+1}-k^n)f(k^{n+1}) \\ &&&= k^n(k-1)f(k^{n+1}) \\ \end{align*}

1. Notice that if \(f(r) = 1/r\) then \(f(r) > f(r+1)\) so we can apply our lemma, ie \begin{align*} &&&2^N(2-1) \frac{1}{2^{N+1}} &\leq & \sum_{r=2^N}^{2^{N+1}-1} \frac1r &\leq&\quad 2^N(2-1) \frac{1}{2^{N}} \\ \Leftrightarrow &&& \frac12 &\leq & \sum_{r=2^N}^{2^{N+1}-1} \frac1r &\leq&\quad 1 \\ \Rightarrow &&& \frac12+\frac12+\cdots+\frac12 &\leq & \underbrace{\sum_{r=2^0}^{2^{0+1}-1} \frac1r+\sum_{r=2^1}^{2^{1+1}-1} \frac1r+\cdots+\sum_{r=2^N}^{2^{N+1}-1} \frac1r}_{N+1 \text{ terms}} &\leq&\quad 1 +1+\cdots+1\\ \Rightarrow &&& \frac{N+1}{2} &\leq & \underbrace{\sum_{r=1}^{2^{N+1}-1} \frac1r}_{N+1 \text{ terms}} &\leq&\quad N+1 \end{align*} Therefore the sum \(\displaystyle \sum_{r=1}^{2^{N+1}-1} \frac1r\) is always greater than \(N+1\) and in particular we can find an upper limit such that it is always bigger than any value, ie it diverges.
2. If \(f(r) = 1/r^3\) then we have \begin{align*} && \sum_{r=2^N}^{2^{N+1}-1} \frac1{r^3} &\leq 2^N(2-1) \frac{1}{2^{3N}} \\ &&&= \frac{1}{4^N} \\ \Rightarrow && \sum_{r=2^0}^{2^{0+1}-1} \frac1{r^3} +\sum_{r=2^1}^{2^{1+1}-1} \frac1{r^3} +\sum_{r=2^N}^{2^{N+1}-1} \frac1{r^3} &\leq 1 + \frac14 + \cdots + \frac1{4^N} \\ \Rightarrow && \sum_{r=1}^{\infty} \frac1{r^3} &\leq 1 + \frac14 + \cdots \\ &&&= \frac{1}{1-\frac14} = \frac43 = 1\tfrac13 \end{align*}
3. To count the number of numbers less than \(1000\) without a \(2\) in their decimal representation we can count the number of \(3\) digit numbers (where \(0\) is an acceptable leading digit) which don't contain a \(2\) and remove \(0\). There are \(9\) choices for each digit, so \(9^3-1\). Notice this is true for \(10^N\) for any \(N\), ie \(S(10^N) = 9^N-1\). Notice also that we can now write: \begin{align*} && \sum_{r=10^N }^{10^{N+1}-1} \frac{1}{r} \mathbb{1}_{r \in S} & < \frac{1}{10^{N+1}}\#\{\text{number not containing a }2\} \\ &&&= \frac{1}{10^{N+1}}((9^{N+1}-1)-(9^N-1)) \\ &&&= \frac{9^N}{10^N}(9-1) \\ &&&= 8 \cdot \left (\frac9{10} \right)^N \\ \\ \Rightarrow && \sum_{r=1}^{\infty} \frac{1}{r} \mathbb{1}_{r \in S} &< 8\left ( 1 + \frac9{10} + \cdots \right) \\ &&&= 8 \frac{1}{1-\frac{9}{10}} = 80 \end{align*}

---

Solution:

1. \(\,\) \begin{align*} && u_{n+2} - u_n &= 1 + \frac{1}{u_{n+1}} - \left (1 + \frac{1}{u_{n-1}} \right) \\ &&&= \frac{1}{1 + \frac1{u_n}} - \frac{1}{1 + \frac{1}{u_{n-2}}} \\ &&&= \frac{u_n}{u_n+1} - \frac{u_{n-2}}{1+u_{n-2}} \\ &&&= \frac{u_n(1+u_{n-2}) - u_{n-2}(1+u_n)}{(1+u_n)(1+u_{n-2})} \\ &&&= \frac{u_n - u_{n-2}}{(1+u_n)(1+u_{n-2})} \end{align*}
2. Claim: \(u_n \in [1,2]\) Proof: (By induction). Note that \(u_1 = 1, u_2 = 2\) so our claim is true for the first few terms. Note that if \(u_n \in [1,2]\), \(\frac{1}{u_n} \in [\tfrac12, 1]\) and \(1+\frac{1}{u_{n}} \in [\tfrac32,2] \subset [1,2]\). Therefore \(u_{n+1} \in [1,2]\). Therefore since \(u_1 \in [1,2]\) and \(u_n \in [1,2] \Rightarrow u_{n+1} \in [1,2]\) \(u_n \in [1,2]\) for all \(n \ge 1\)
3. First notice that \(u_3 = \frac32 > u_1\) and therefore by the recursion we found in the first part, \(u_{2n+1}-u_{2n-1} > 0\) so \(u_{2k+1}\) is increasing and bounded, and so by our theorem converges to a limit. Suppose this limit is \(L\), then we must have \(L = 1 + \frac1{L} \Rightarrow L^2 - L - 1 = 0 \Rightarrow L = \frac{1+\sqrt5}{2}\) since it must be in \([1,2]\). Similarly, not that \(u_4 = \frac{5}{3} < u_2\) and so \(u_{2k+2} - u_{2k} < 0\) and \(-u_{2k}\) is increasing and bounded above. Therefore it tends to a limit (and so does \(u_{2k}\)). By the same reasoning as before, it's the same limit, \(\frac{1+\sqrt5}{2}\) and therefore the sequence converges. If \(u_1 = 3, u_2 = \frac43 \in [1,2]\) so we have our sequence being bounded and all the same logic will follow through.

---

Solution:

1. Since \(k \geq 1\) we have \(n^{k+1} > n^k\) and \((k+1) > k\), therefore \((k+1)n^{k+1} >kn^k \Rightarrow \frac{1}{(k+1)n^{k+1}} < \frac{1}{kn^k}\) \begin{align*} && \ln \left ( 1 + \frac1n \right) &= \frac1n -\frac{1}{2n^2} + \frac{1}{3n^3} - \frac{1}{4n^4} + \cdots \\ &&&= \frac1n - \underbrace{\left (\frac{1}{2n^2}-\frac{1}{3n^3} \right)}_{>0}- \underbrace{\left (\frac{1}{4n^4}-\frac{1}{5n^5} \right)}_{>0} - \cdot \\ &&&< \frac1n \\ \\ \Rightarrow && n \ln \left ( 1 + \frac1n \right) &< 1 \\ \Rightarrow && \ln \left ( \left ( 1 + \frac1n \right)^n \right) &< 1 \\ \Rightarrow && \left ( 1 + \frac1n \right)^n &< e \end{align*}
2. \(\,\) \begin{align*} &&\ln \left(\frac{2y+1}{2y-1}\right) &= \ln \left (1 + \frac{1}{2y} \right)-\ln \left (1 - \frac{1}{2y} \right) \\ &&&= \frac{1}{2y} - \frac{1}{2(2y)^2} + \frac{1}{3(2y)^3} - \cdots - \left (-\frac{1}{2y} - \frac{1}{2(2y)^2} - \frac{1}{3(2y)^3} - \cdots \right) \\ &&&= \frac{1}{y} + \frac{2}{3(2y)^3} + \frac{2}{5(2y)^5} \\ &&&= \sum_{r=1}^{\infty} \frac{2}{(2r-1)(2y)^{2r-1}} \\ &&&> \frac1y \\ \\ \Rightarrow && \ln \left (1 + \frac{1}{y-\frac12} \right) &> \frac{1}{y} \\\Rightarrow && \ln \left (1 + \frac{1}{n} \right) &> \frac{1}{n+\frac12} \\ \Rightarrow &&(n+\tfrac12) \ln \left (1 + \frac{1}{n} \right) &> 1\\ \Rightarrow && \ln \left ( \left (1 + \frac{1}{n} \right)^{n+\tfrac12} \right) &> 1\\ \Rightarrow && \left (1 + \frac{1}{n} \right)^{n+\tfrac12} & > e \end{align*}

\item Since \(\left (1 + \frac1n \right)^n\) is both bounded above, and increasing, it must tend to some limit \(L\). \begin{align*} && \lim_{n \to \infty} \left (1 + \frac1n \right)^n && \leq e &\leq & \lim_{n \to \infty} \left (1 + \frac1n \right)^{n+\frac12} \\ \Rightarrow && \lim_{n \to \infty} \left (1 + \frac1n \right)^n && \leq e &\leq & \lim_{n \to \infty} \left (1 + \frac1n \right)^{n} \lim_{n \to \infty} \sqrt{1 + \frac1n} \\ \Rightarrow && \lim_{n \to \infty} \left (1 + \frac1n \right)^n && \leq e &\leq & \lim_{n \to \infty} \left (1 + \frac1n \right)^{n} \\ \end{align*} And therefore equality must hold.

---

Solution:

1. \begin{align*} (1-x)&(1+x)(1+x^2)(1+x^4) \cdots (1+x^{2^n}) \\ &= (1-x^2)(1+x^2)(1+x^4) \cdots (1+x^{2^n}) \\ &= (1-x^4)(1+x^4) \cdots (1+x^{2^n}) \\ &= 1-x^{2^{n+1}} \\ \end{align*} Therefore, \begin{align*} && \frac{1}{1-x} - \frac{x^{2^{n+1}}}{1-x} &= (1+x)(1+x^2)\cdots(1+x^{2^n}) \\ \Rightarrow && \frac{1}{1-x} &=(1+x)(1+x^2)\cdots(1+x^{2^n})+ \frac{x^{2^{n+1}}}{1-x} \\ \Rightarrow && -\ln (1-x) &= \sum_{r=0}^{\infty} \ln (1+x^{2^r}) \\ \Rightarrow && \ln(1-x) &= - \sum_{r=0}^{\infty} \ln (1+x^{2^r}) \\ \underbrace{\Rightarrow}_{\frac{\d}{\d x}} && \frac{1}{1-x} &= \sum_{r=0}^{\infty} \frac{2^r x^{2^r-1}}{1+x^{2^r}} \\ &&&= \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \cdots \end{align*}
2. Consider \begin{align*}(1+x+x^2)&(1-x+x^2)(1-x^2+x^4)\cdots(1-x^{2^n}+x^{2^{n+1}}) \\ &= (1+x^2 + x^4)(1-x^2+x^4) \cdots (1-x^{2^n}+x^{2^{n+1}}) \\ &= (1-x^{2^{n+1}}+x^{2^{n+2}}) \\ \end{align*} Therefore, \begin{align*} && \frac{1}{1+x+x^2} &= (1-x+x^2)(1-x^2+x^4)\cdots(1-x^{2^n}+x^{2^{n+1}}) + \frac{x^{2^{n+1}}}{1+x+x^2} -\frac{x^{2^{n+2}}}{1+x+x^2} \\ \Rightarrow && -\ln(1+x+x^2) &= \sum_{r=0}^\infty \ln (1 - x^{2^r}+x^{2^{r+1}}) \\ \underbrace{\Rightarrow}_{\frac{\d}{\d x}} && -\frac{1+2x}{1+x+x^2} &= \sum_{r=0}^{\infty} \frac{ -2^r x^{2^r-1}+2^{r+1}x^{2^{r+1}-1}}{1 - x^{2^r}+x^{2^{r+1}}} \\ &&&= \frac{-1+2x}{1-x+x^2}+\frac{-2x+4x^3}{1-x^2+x^4} + \frac{-4x^3+8x^7}{1-x^4+x^8} + \cdots \end{align*} Which is the desired result when we multiply both sides by \(-1\)

---

Solution:

1. Notice that \(1 \cdot 3 \cdot 5 \cdot 7 \cdot (2n-1) = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdots \cdots 2n}{2 \cdot 4 \cdot 6 \cdots 2n} = \frac{(2n)!}{2^n \cdot n!}\) as required. When \(|4x| < 1\) or \(|x|<\frac14\) we can apply the generalised binomial theorem to see that: \begin{align*} \frac{1}{\sqrt{1-4x}} &= (1-4x)^{-\frac12} \\ &= 1+\sum_{n=1}^\infty \frac{-\frac12 \cdot \left ( -\frac32\right)\cdots \left ( -\frac{2n-1}2\right)}{n!} (-4x)^n \\ &= 1+\sum_{n=1}^{\infty} (-1)^n\frac{(2n)!}{(n!)^2 2^{2n}} (-4)^n x^n \\ &= 1+\sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2 } x^n \\ \end{align*}
2. Differentiating we obtain \begin{align*} && 2(1-4x)^{-\frac32} &= \sum_{n=1}^\infty \frac{(2n)!}{n!(n-1)!} x^{n-1} \\ \Rightarrow &&\sum_{n=1}^\infty \frac{(2n)!}{n!(n-1)!} \left (\frac{6}{25} \right)^{n} &= \frac{6}{25} \cdot 2\left(1- 4 \frac{6}{25}\right)^{-\frac32} \\ &&&= \frac{12}{25} \left (\frac{1}{25} \right)^{-\frac32} \\ &&&= \frac{12}{25} \cdot 125 = 60 \end{align*}
3. By integrating, we obtain \begin{align*} && \int_{t=0}^{t=x} \frac{1}{\sqrt{1-4t}} \d t &= \int_{t=0}^{t=x} \left (1+\sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2 } t^n \right) \d t \\ \Rightarrow && \left [ -\frac12 \sqrt{1-4t}\right]_0^x &= x + \sum_{n=1}^{\infty} \frac{(2n)!}{n!(n+1)! } x^{n+1} \\ \Rightarrow && \frac12 - \frac12\sqrt{1-4x} - x &= \sum_{n=1}^{\infty} \frac{(2n)!}{n!(n+1)! } x^{n+1} \\ \\ \Rightarrow && 9\sum_{n=1}^{\infty} \frac{(2n)!}{n!(n+1)! } \left ( \frac{2}{9}\right)^{n+1} &=9 \cdot\left( \frac12 - \frac12 \sqrt{1-4\cdot \frac{2}{9}} - \frac29\right )\\ &&&= 9 \cdot \left (\frac12 - \frac{1}{6} - \frac{2}{9} \right) \\ &&&= 1 \end{align*}

---

Solution:

1. Suppose \(x_n> 3\) then \begin{align*} && x_{n+1} &= \frac{x_n^2+9-3}{5} \\ &&& \geq \frac{2\sqrt{x_n^2 \cdot 9} - 3}{5} \\ &&&= \frac{6x_n -3}{5} = x_n + \frac{x_n-3}{5} \\ &&&> x_n > 3 \end{align*} Therefore if \(x_i > 3 \Rightarrow x_{i+1} > x_i\) and \(x_{i+1} > 3\) so by induction \(x_n\) strictly increasing for all \(n\).
2. Suppose \(2 < y_n < 3\) then \begin{align*} && y_{n+1} &= 5 - \frac6{y_n} \\ &&&< 5 - \frac63 = 3 \\ \\ && y_{n+1} &= 5 - \frac4{y_n} - \frac{2}{y_n} \\ \\ &&&= y_n + 5 - \frac{2}{y_n} - \left ( y_n + \frac4{y_n} \right) \\ &&&\geq y_n + 5 - \frac{2}{y_n} - 2\sqrt{y_n \frac{4}{y_n}} \\ &&&= y_n + 1 - \frac{2}{y_n} \\ &&&> y_n \end{align*} Therefore if \(y_n \in (2,3)\) we have \(y_{n+1} \in ( y_n, 3)\) and so \(y_n\) is strictly increasing and bounded.

---

Solution: Claim: \(x \in \mathbb{R}\setminus \mathbb{Q} \Rightarrow x^{1/3} \in \mathbb{R} \setminus\mathbb{Q}\) Proof: We will prove the contrapositive, since \(x^{1/3} = p/q\) but then \(x = p^3/q^3 \in \mathbb{Q}\), therefore we're done. Claim: \(u_n = 5^{1/(3^n)}\) is irrational for \(n \geq 1\) Proof: We are assuming the base case, but then \(u_{n+1} = \sqrt[3]{u_n}\) which is clearly irrational by our first lemma, so we're done. Note that \(u_n \to 1\) and so \((m-1)+u_n \to m\) for any integer \(m\).