Verified answer

For fixed x, let a_n = e^{nx}. Then (a_n)^{1/n} = e^x. By the ratio test the series ∑ a_n converges if lim (a_n)^{1/n} < 1 and diverges if lim (a_n)^{1/n} > 1, i.e., ∑ a_n converges if e^x < 1 and diverges if e^x > 1. e^x < 1 if and only if x < 0, thus ∑ a_n converges if x is negative. Note that if x = 0, then the series becomes ∑ 1, which is divergent. Therefore ∑ e^{nx} converges for all negative x.