Which of the following sequences are unbounded?

$$\begin{gathered} \text{ (a) }\because \mathrm{S}={\left(1+\frac{1}{\mathrm{n}}\right)}^{\mathrm{n}} \\ \mathrm{S}=1{+}^{\mathrm{n}}{\mathrm{C}}_1\left(\frac{1}{\mathrm{n}}\right){+}^{\mathrm{n}}{\mathrm{C}}_2{\left(\frac{1}{\mathrm{n}}\right)}^2{+}^{\mathrm{n}}{\mathrm{C}}_3{\left(\frac{1}{\mathrm{n}}\right)}^3+\dots {+}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}{\left(\frac{1}{\mathrm{n}}\right)}^{\mathrm{n}} \\ \mathrm{S}=1+\mathrm{n}\cdot \frac{1}{\mathrm{n}}+\frac{\mathrm{n}(\mathrm{n}-1)}{2!}\left(\frac{1}{{\mathrm{n}}^2}\right)+\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{3!}\left(\frac{1}{{\mathrm{n}}^3}\right)+\dots +\frac{\mathrm{n}(\mathrm{n}-1)\dots 1}{\mathrm{n}!}\left(\frac{1}{{\mathrm{n}}^{\mathrm{n}}}\right) \\ \mathrm{S}=1+1+\frac{1}{2!}\left(1-\frac{1}{\mathrm{n}}\right)+\frac{1}{3!}\left(1-\frac{1}{\mathrm{n}}\right)\left(1-\frac{2}{\mathrm{n}}\right)+\dots +\frac{1}{\mathrm{n}!}\left(1-\frac{1}{\mathrm{n}}\right)\left(1-\frac{2}{\mathrm{n}}\right)\cdots \left(1-\frac{\mathrm{n}-1}{\mathrm{n}}\right) \\ \mathrm{S}<1+1+\frac{1}{2!}+\frac{1}{3!}+\dots +\frac{1}{\mathrm{n}!} \\ \begin{array}{r}\mathrm{S}<1+1+\frac{1}{1\cdot 2}+\frac{1}{1\cdot 2\cdot 3}+\dots +\frac{1}{1\cdot 2\cdot 3\dots \mathrm{n}}\\ \mathrm{S}<1+1+\frac{1}{2}+\frac{1}{2\cdot 2}+\dots +\frac{1}{2\cdot 2\dots 2}\end{array} \\ \mathrm{s}<1+\frac{1-\frac{1}{2^{\mathrm{n}}}}{1-\frac{1}{2}}\Rightarrow \mathrm{s}<1+2\left(1-\frac{1}{2^{\mathrm{n}}}\right) \\ \mathrm{S}<3-\frac{1}{2^{\mathrm{n}-1}}\therefore \mathrm{S}<3,\mathrm{\forall }\mathrm{n} \\ \mathrm{Also}, \mathrm{S}=1+1+\frac{1}{2!}\left(1-\frac{1}{\mathrm{n}}\right)+\frac{1}{3!}\left(1-\frac{1}{\mathrm{n}}\right)\left(1-\frac{2}{\mathrm{n}}\right)+\dots +\frac{1}{\mathrm{n}!}\left(1-\frac{1}{\mathrm{n}}\right)\left(1-\frac{2}{\mathrm{n}}\right)\cdots \left(1-\frac{\mathrm{n}-1}{\mathrm{n}}\right) \\ \mathrm{S}>1+1;\mathrm{S}>2 \\ \therefore \mathrm{S}\text{ is bounded. } \end{gathered}$$
$$\begin{gathered} \left(\mathrm{b}\right) \because {\mathrm{a}}_{\mathrm{n}}=\frac{2\mathrm{n}+1}{\mathrm{n}+2} \\ \mathrm{For} {\mathrm{n}}_1=1,{\mathrm{a}}_1=\frac{3}{3}=1, \\ \mathrm{for} \mathrm{n}=2,{\mathrm{a}}_2=\frac{5}{4}=1.25 \\ :\; :\; : \\ \mathrm{Now}, {\mathrm{a}}_{\mathrm{n}+1}-{\mathrm{a}}_{\mathrm{n}}>0\Rightarrow {\mathrm{a}}_{\mathrm{n}+1}>{\mathrm{a}}_{\mathrm{n}} \\ \therefore {\mathrm{a}}_{\mathrm{n}} \mathrm{represents} \mathrm{the} \mathrm{increasing} \mathrm{sequence} \\ \underset{\mathrm{n}\to \mathrm{\infty }}{lim} {\mathrm{a}}_{\mathrm{n}}=\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \frac{2\mathrm{n}+1}{\mathrm{n}+2}=\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \frac{\mathrm{n}\left(2+\frac{1}{\mathrm{n}}\right)}{\mathrm{n}\left(1+\frac{2}{\mathrm{n}}\right)}=\frac{2}{1}=2 \\ \therefore \left\{{\mathrm{a}}_{\mathrm{n}}\right\} \mathrm{is} \mathrm{bounded} \mathrm{sequence}. \end{gathered}$$
$$\begin{gathered} \text{ (c) }\because ,{\mathrm{a}}_{\mathrm{n}}={\left(1+\frac{1}{\mathrm{n}}\right)}^{{\mathrm{n}}^2} \\ \text{ For }\mathrm{n}=1,{\mathrm{a}}_1=2\text{, } \\ \mathrm{for} \mathrm{n}=2,{\mathrm{a}}_2={\left(1+\frac{1}{2}\right)}^4={\left(\frac{3}{2}\right)}^4=\frac{3^4}{2^4}=\frac{81}{16}=5.06 \left[\mathrm{approximate}\right] \\ \underset{\mathrm{n}\to \mathrm{\infty }}{lim} {\left(1+\frac{1}{\mathrm{n}}\right)}^{{\mathrm{n}}^2}={\mathrm{e}}^{\underset{\mathrm{n}\to -\mathrm{n}}{lim} \frac{1}{\mathrm{n}}\times {\mathrm{n}}^2}={\mathrm{e}}^{\mathrm{n}\to --\mathrm{n}}={\mathrm{e}}^{\mathrm{\infty }}=\mathrm{\infty } \\ \therefore \left\{{\mathrm{a}}_{\mathrm{n}}\right\}\text{ represents unbounded sequence. } \\ \text{ (d) }\because {\mathrm{a}}_{\mathrm{n}}=\tan \mathrm{n} \\ {\mathrm{a}}_{\mathrm{n}}=\mathrm{n}+\frac{{\mathrm{n}}^3}{3}+\frac{2}{15}{\mathrm{n}}^5+\dots +\mathrm{\infty } \end{gathered}$$
$$\begin{gathered} \mathrm{and} \mathrm{we} \mathrm{know} \mathrm{that} -\mathrm{\infty }<\tan \mathrm{n}<\mathrm{\infty } \\ \mathrm{So}, \left\{{\mathrm{a}}_{\mathrm{n}}\right\} \mathrm{is} \mathrm{unbounded} \mathrm{sequence}. \end{gathered}$$