Exponential Series Convergence

Note that $e^{x} - 1 = x+ \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots.$ Since $\sum_{k=1}^\infty x_k$ converges, $\lim_{k\rightarrow\infty}x_k = 0.$ Thus, for sufficiently large $k\ge K,$ $0< x_k < 1,$ implying that $x_k>x_k^2 > x_k^3 > \cdots.$ So, for $k \ge K,$ $e^{x_k} -1 = x_k+ \frac{x_k^2}{2!} + \frac{x_k^3}{3!} + \cdots < x_k+ \frac{x_k}{2!} + \frac{x_k}{3!} + \cdots = x_k\left(1 + \frac{1}{2!} + \frac{1}{3!} + \cdots\right) = x_k(e-1).$

The sum $\sum_{k=1}^{K-1} \left(e^{x_k}-1\right)$ is finitely many finite terms; then, $\sum_{k=K}^{\infty} \left(e^{x_k}-1\right) < (e-1)\cdot \sum_{k=K}^{\infty} x_k,$ and the right hand side converges since $\sum_{k=1}^\infty x_k$ converges.

let consider $f(y) = e^y$ over the interval $[0, x]$ for $x$ positive.

By the Mean value theorem , it exists c in $[0, x]$ such: $f'(c) = \dfrac {f(x) - f(0)}{x - 0}$ This yields: $\dfrac {e^x - 1}{x}\ = e^c < e^x$

So that for $x$ in $[0,1]$ ${e^x - 1} < e \times x$

Since $\sum_{k=1}^\infty x_k$ converges, $\lim_{k\rightarrow\infty}x_k = 0.$

Thus, for sufficiently large $k\ge K,$ $0< x_k < 1,$ implying that for $k \ge K,$ $e^{x_k} -1 < x_k \times e.$

by the Comparison Test of convergence $\sum_{k=1}^{\infty}\left(e^{x_k}-1\right)$ est convergente.