Show that $S _ { N }$ converges uniformly to $S$ on $\mathbf { R } _ { + }$ when $N \rightarrow + \infty$ and that $$\left\| S _ { N } - S \right\| _ { \infty , \mathbf { R } _ { + } } \leq \frac { M \mathrm { e } ^ { 2 M } } { 2 ^ { N + 1 } }$$ where $S _ { N } ( x ) = S _ { 0 } \mathrm { e } ^ { y _ { N } ( x ) } = \frac { 1 } { 2 } \exp \left( \sum _ { n = 0 } ^ { N } a _ { n } \mathrm { e } ^ { - \lambda _ { n } x } \right)$, $S(x) = S_0 e^{y(x)} = \frac{1}{2} e^{y(x)}$, and $\left\| y_N - y \right\|_{\infty, \mathbf{R}_+} \leq \frac{M}{2^N}$.
Show that $S _ { N }$ converges uniformly to $S$ on $\mathbf { R } _ { + }$ when $N \rightarrow + \infty$ and that
$$\left\| S _ { N } - S \right\| _ { \infty , \mathbf { R } _ { + } } \leq \frac { M \mathrm { e } ^ { 2 M } } { 2 ^ { N + 1 } }$$
where $S _ { N } ( x ) = S _ { 0 } \mathrm { e } ^ { y _ { N } ( x ) } = \frac { 1 } { 2 } \exp \left( \sum _ { n = 0 } ^ { N } a _ { n } \mathrm { e } ^ { - \lambda _ { n } x } \right)$, $S(x) = S_0 e^{y(x)} = \frac{1}{2} e^{y(x)}$, and $\left\| y_N - y \right\|_{\infty, \mathbf{R}_+} \leq \frac{M}{2^N}$.