Answer
We show that $\dfrac{{Vt}}{L}\left( {1 - \dfrac{R}{{2L}}t} \right) \le I\left( t \right) \le \dfrac{{Vt}}{L}$ ${\ \ \ }$ (for all $t$)
Work Step by Step
From Exercise 71 we know that
$I\left( t \right) = \left( {\dfrac{V}{R}} \right)\left( {1 - {{\rm{e}}^{ - Rt/L}}} \right)$
From Table 2, we get
${{\rm{e}}^x} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{x^n}}}{{n!}} = 1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$.
Substituting $ - Rt/L$ for $x$ in the series above gives
${{\rm{e}}^{ - Rt/L}} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - Rt/L} \right)}^n}}}{{n!}} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{{\left( {Rt/L} \right)}^n}}}{{n!}} = 1 - \dfrac{{Rt}}{L} + \dfrac{{{{\left( {Rt/L} \right)}^2}}}{{2!}} - \dfrac{{{{\left( {Rt/L} \right)}^3}}}{{3!}} + \dfrac{{{{\left( {Rt/L} \right)}^4}}}{{4!}} - \cdot\cdot\cdot$
Thus,
$I\left( t \right) = \left( {\dfrac{V}{R}} \right)\left( {1 - {{\rm{e}}^{ - Rt/L}}} \right)$
$I\left( t \right) = \left( {\dfrac{V}{R}} \right)\left( {\dfrac{{Rt}}{L} - \dfrac{{{{\left( {Rt/L} \right)}^2}}}{{2!}} + \dfrac{{{{\left( {Rt/L} \right)}^3}}}{{3!}} - \dfrac{{{{\left( {Rt/L} \right)}^4}}}{{4!}} + \cdot\cdot\cdot} \right)$
$I\left( t \right) = \left( {\dfrac{V}{R}} \right)\mathop \sum \limits_{n = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^{n - 1}}{{\left( {Rt/L} \right)}^n}}}{{n!}}$
$I\left( t \right) = \mathop \sum \limits_{n = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^{n - 1}}V{R^{n - 1}}{t^n}}}{{{L^n}n!}} = \dfrac{{Vt}}{L} - \dfrac{{VR{{\left( {t/L} \right)}^2}}}{{2!}} + \dfrac{{V{R^2}{{\left( {t/L} \right)}^3}}}{{3!}} - \dfrac{{V{R^3}{{\left( {t/L} \right)}^4}}}{{4!}} + \cdot\cdot\cdot$
We have the positive terms given by ${b_n} = \dfrac{{V{R^{n - 1}}{t^n}}}{{{L^n}n!}}$. Since $\left\{ {{b_n}} \right\}$ is positive sequence that is decreasing and converges to $0$, we can apply Theorem 2 in Section 11.4, that is
$0 \le I\left( t \right) \le {b_1}$ ${\ \ \ \ }$ and ${\ \ \ \ }$ ${I_{2N}} \le I\left( t \right) \le {I_{2N + 1}}$
Since the sequence is decreasing, so
${I_{2N}} \le I\left( t \right) \le {I_{2N + 1}} \le {b_1}$
Setting $N=1$, we obtain
${I_2} \le I\left( t \right) \le {b_1}$
Getting ${I_2}$ from the expansion in equation (1) and substituting it in the inequality above gives
$\dfrac{{Vt}}{L} - \dfrac{{VR{{\left( {t/L} \right)}^2}}}{{2!}} \le I\left( t \right) \le \dfrac{{Vt}}{L}$
$\dfrac{{Vt}}{L}\left( {1 - \dfrac{R}{{2L}}t} \right) \le I\left( t \right) \le \dfrac{{Vt}}{L}$
Hence, $\dfrac{{Vt}}{L}\left( {1 - \dfrac{R}{{2L}}t} \right) \le I\left( t \right) \le \dfrac{{Vt}}{L}$ ${\ \ \ }$ (for all $t$)
