Prove that the series $\sum_{n=1}^\infty e^{-nP(n)/Q(n)}$ converges. Let $P(x)$ and $Q(x)$ be polynomials of the same degree $d>0$. Suppose that the coefficients of the polynomials are nonnegative real numbers. Prove that the series

$$\sum_{n=1}^\infty e^{-nP(n)/Q(n)}$$

converges.
I'm pretty stuck here, even in terms of which direction to go. I considered the ratio test, but the algebra is incredibly messy, as would any sort of polynomial long division be.
Any help would be appreciated!
 A: Let
$$0<\alpha =\lim_{n\to\infty}{P(n)\over Q(n)}$$
which exists and is positive by your assumptions.
Then by the limit comparison test with $b_n= e^{-\alpha n}$ we see your series converges iff
$$\sum_{n=1}^\infty e^{-\alpha n}$$
does. But of course this series converges by integral test.
A: Apply the nth root test! After writing the limit, and bringing the power of $\frac{1}{n}$ inside, the problem turns into a question of when $\lim \limits_{n \to \infty}| \exp\big(-\frac{P(n)}{Q(n)} \big)| <1$. Since the argument of the absolute value is always positive, the absolute value can be dropped. It might be worth rewriting the problem as $\lim \limits_{n \to \infty} \frac{1}{exp \big( \frac{P(n)}{Q(n)} \big)}$ . Since by the way you defined it, $\frac{P(n)}{Q(n)}$ will tend to the ratio of the coefficients and that ratio will be positive. Picking the coefficients as you please, $\lim \limits_{n \to \infty} \frac{P(n)}{Q(n)}$ can be made to be arbitrarily large, or arbitrarily close to zero. In any case, $\lim_\limits{n \to \infty}| \exp\big(-\frac{P(n)}{Q(n)} \big)|$ will be between, but not including $0$ and $1$. This satisfies convergence by the root test.
