2017.3.1 Question 1

1. We have

   $$\begin{array}{llll}\hfill \text{RHS}& =\frac{r+1}{r}\left(\frac{1}{\left(\genfrac{}{}{0.0pt}{}{n+r-1}{r}\right)}-\frac{1}{\left(\genfrac{}{}{0.0pt}{}{n+r}{r}\right)}\right)& \hfill & \\ \hfill & =\frac{r+1}{r}\left(\frac{r!(n-1)!}{(n+r-1)!}-\frac{r!n!}{(n+r)!}\right)& \hfill & \\ \hfill & =\frac{r+1}{r}\left(\frac{r!(n-1)!(n+r)}{(n+r)!}-\frac{r!(n-1)!n}{(n+r)!}\right)& \hfill & \\ \hfill & =\frac{r+1}{r}\cdot \frac{r!(n-1)!(n+r)-r!(n-1)!n}{(n+r)!}& \hfill & \\ \hfill & =\frac{r+1}{r}\cdot \frac{r!(n-1)!r}{(n+r)!}& \hfill & \\ \hfill & =\frac{(r+1)!(n-1)!}{(n+r)!}& \hfill & \\ \hfill & =\left(\genfrac{}{}{0.0pt}{}{n+r}{r+1}\right)& \hfill & \\ \hfill & =\text{LHS}& \hfill & \end{array}$$

   as desired.

   Therefore,

   $$\begin{array}{llll}\hfill \sum _{n=1}^{+\infty }\frac{1}{\left(\genfrac{}{}{0.0pt}{}{n+r}{r+1}\right)}& =\sum _{n=1}^{+\infty }\frac{r+1}{r}\left(\frac{1}{\left(\genfrac{}{}{0.0pt}{}{n+r-1}{r}\right)}-\frac{1}{\left(\genfrac{}{}{0.0pt}{}{n+r}{r}\right)}\right)& \hfill & \\ \hfill & =\frac{r+1}{r}\sum _{n=1}^{+\infty }\left(\frac{1}{\left(\genfrac{}{}{0.0pt}{}{n+r-1}{r}\right)}-\frac{1}{\left(\genfrac{}{}{0.0pt}{}{n+r}{r}\right)}\right)& \hfill & \\ \hfill & =\frac{r+1}{r}\left[\sum _{n=0}^{+\infty }\frac{1}{\left(\genfrac{}{}{0.0pt}{}{n+r}{r}\right)}-\sum _{n=1}^{+\infty }\frac{1}{\left(\genfrac{}{}{0.0pt}{}{n+r}{r}\right)}\right]& \hfill & \\ \hfill & =\frac{r+1}{r}\frac{1}{\left(\genfrac{}{}{0.0pt}{}{0+r}{r}\right)}& \hfill & \\ \hfill & =\frac{r+1}{r},& \hfill & \end{array}$$

   assuming the sum converges.

   When $r=2$, we have

   $$\sum _{n=1}^{+\infty }\frac{1}{\left(\genfrac{}{}{0.0pt}{}{n+2}{3}\right)}=\frac{3}{2}.$$

   When $n=1$, $\frac{1}{\left(\genfrac{}{}{0.0pt}{}{1+2}{3}\right)}=\frac{1}{1}=1$.

   Therefore,

   $$\sum _{n=2}^{+\infty }\frac{1}{\left(\genfrac{}{}{0.0pt}{}{n+2}{3}\right)}=\frac{1}{2}$$

   as desired.

2. Notice that

   $$\begin{array}{llll}\hfill \frac{3!}{n^3}<\frac{1}{\left(\genfrac{}{}{0.0pt}{}{n+1}{3}\right)}& ⟺\frac{3!}{n^3}<\frac{3!}{(n+1)n(n-1)}& \hfill & \\ \hfill & ⟺n^3>(n+1)n(n-1)& \hfill & \\ \hfill & ⟺n^3>n(n^2-1)& \hfill & \\ \hfill & ⟺n^3>n^3-n& \hfill & \\ \hfill & ⟺n>0,& \hfill & \end{array}$$

   which is true.

   Also, notice that

   $$\begin{array}{llll}\hfill \frac{20}{\left(\genfrac{}{}{0.0pt}{}{n+1}{3}\right)}-\frac{1}{\left(\genfrac{}{}{0.0pt}{}{n+2}{5}\right)}<\frac{5!}{n^3}& ⟺\frac{5!}{(n+1)(n)(n-1)}-\frac{5!}{(n+2)(n+1)(n)(n-1)(n-2)}<\frac{5!}{n^3}& \hfill & \\ \hfill & ⟺\frac{(n+2)(n-2)-1}{(n+2)(n+1)(n)(n-1)(n-2)}<\frac{1}{n^3}& \hfill & \\ \hfill & ⟺(n^2-5)n^3<(n^2-4)(n^2-1)n& \hfill & \\ \hfill & ⟺n^5-5n^3<n^5-5n^3+4n& \hfill & \\ \hfill & ⟺4n>0,& \hfill & \end{array}$$

   which is true.

   Therefore, we have that

   $$\begin{array}{llll}\hfill \sum _{n=3}^{+\infty }\frac{3!}{n^3}& <\sum _{n=3}^{+\infty }\frac{1}{\left(\genfrac{}{}{0.0pt}{}{n+1}{3}\right)}& \hfill & \\ \hfill & =\sum _{n=2}^{+\infty }\frac{1}{\left(\genfrac{}{}{0.0pt}{}{n+2}{3}\right)}& \hfill & \\ \hfill & =\frac{1}{2},& \hfill & \end{array}$$

   and therefore ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=3}^{+\infty }\frac{1}{n^3}<\frac{1}{12}$, and ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{+\infty }\frac{1}{n^3}<1+\frac{1}{8}+\frac{1}{12}=\frac{29}{24}=\frac{116}{96}.$

   On the other hand, we have

   $$\begin{array}{llll}\hfill \sum _{n=3}^{+\infty }\frac{5!}{n^3}& <\sum _{n=3}^{+\infty }\left[\frac{20}{\left(\genfrac{}{}{0.0pt}{}{n+1}{3}\right)}-\frac{1}{\left(\genfrac{}{}{0.0pt}{}{n+2}{5}\right)}\right]& \hfill & \\ \hfill & =20\sum _{n=2}^{+\infty }\frac{1}{\left(\genfrac{}{}{0.0pt}{}{n+2}{3}\right)}-\sum _{n=1}^{+\infty }\frac{1}{\left(\genfrac{}{}{0.0pt}{}{n+4}{5}\right)}& \hfill & \\ \hfill & =20\cdot \frac{1}{2}-\frac{5}{4}& \hfill & \\ \hfill & =10-\frac{5}{4}& \hfill & \\ \hfill & =\frac{35}{4},& \hfill & \end{array}$$

   and therefore ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=3}^{+\infty }\frac{1}{n^3}>\frac{7}{96}$, and ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{+\infty }\frac{1}{n^3}>1+\frac{1}{8}+\frac{7}{96}=\frac{115}{96}$.

   Hence,

   $$\frac{115}{96}<\sum _{n=1}^{+\infty }\frac{1}{n^3}<\frac{116}{96}$$

   as desired.