Sum Of Minuscule Differences

Relevant wiki: Convergence of Sequences

Because of non-negativity and convergence of the series, it is clear that there exists a finite index $m \in \mathbb{N}$ , such that $U_n \leq 1$ for all $n\geq m$ . Hence, for any $n\geq m$ , we have $e^{U_n}\leq 1+2U_n$ Thus, $\sum_{n=1}^{\infty}\big(e^{U_n}-1 \big)= \sum_{n=1}^{m-1}\big(e^{U_n}-1 \big)+\sum_{n=m}^{\infty}\big(e^{U_n}-1 \big) \leq \sum_{n=1}^{m-1}\big(e^{U_n}-1 \big) +2 \sum_{n=1}^{\infty} U_n \stackrel{(a)}{<}\infty$ where the inequality (a) follows from the fact that the first term is a finite term and the second term is a convergent series and we have used non-negativity of $U_n$ 's. Hence the given monotone non-decreasing series is bounded above, thus convergent. $\blacksquare$

Given that $\sum_{n=1}^{\infty} U_n$ converges, one can conclude that there must be some $N\geq 1$ such that $0<U_n<1$ if $n\geq N$ . Since $e^x$ is differentiable on the half-open interval $[0,1)$ , from the mean value theorem, one has

$\dfrac{e^x-e^0}{x}=e^c$

for some $0\leq c <1$ . $e^x$ is increasing so one has $e^x-1<e\:x$ for all $x\in [0,1)$ . Thus one sees that

$\sum_{n=N}^\infty (e^{U_n}-1)<e\sum_{n=N}^{\infty} U_n<\infty$ .