Problems

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*} && \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*}