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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!

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    Try applying the euclidean algorithim to $P(n)/Q(n)$ to get that $P(n)/Q(n) = c+\ell(n)$ where $c>0$ and $\ell(n)\in O(n^{-1})$. – 2017-01-16
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    if they have same degree, then their ratio approaches to ratio of their coefficients which corresponds to highest degree in general, so you will have sum of exponential with negative power – 2017-01-16
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    Try the root test? – 2017-01-16

2 Answers 2

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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.

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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.

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    Actually that series is a geometric series. – 2017-01-16
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    @zhw. yes this is another way to see it. – 2017-01-16