β>1时，级数$$\sum_{n=1}^\infty \frac{1}{n^\alpha \ln(1+n^\beta)}$$收敛的α>0取值判定

Alright, let's work through this problem step by step. First, we can simplify the general term using asymptotic equivalence—this is a go-to trick for figuring out series convergence.

Step 1: Simplify the denominator with asymptotic equivalence #

Since β>1, as $n \to \infty$, $n^\beta$ blows up to infinity. Let's rewrite the logarithm term to see its behavior:
$$\ln(1+n^\beta) = \ln\left(n^\beta \left(1 + \frac{1}{n^\beta}\right)\right) = \beta \ln n + \ln\left(1 + \frac{1}{n^\beta}\right)$$
As $n$ gets really large, $\ln\left(1 + \frac{1}{n^\beta}\right)$ approaches 0 (since $\frac{1}{n^\beta}$ goes to 0). So we can say:
$$\ln(1+n^\beta) \sim \beta \ln n$$
This means the general term of our series is asymptotically equivalent to:
$$\frac{1}{n^\alpha \ln(1+n^\beta)} \sim \frac{1}{\beta n^\alpha \ln n}$$
Since β is just a positive constant, the convergence of our original series is exactly the same as the convergence of $\sum_{n=2}^\infty \frac{1}{n^\alpha \ln n}$ (we start at n=2 to avoid ln1=0, which doesn't affect the overall convergence).

Step 2: Analyze convergence based on α #

We'll break this into three cases for α>0:

Case 1: α>1 #

For $n \geq 2$, $\ln n > 1$, so:
$$\frac{1}{n^\alpha \ln n} < \frac{1}{n^\alpha}$$
The series $\sum_{n=1}^\infty \frac{1}{n^\alpha}$ is a p-series with p=α>1, which we know converges. By the comparison test, our simplified series converges—so the original series does too.

Case 2: α=1 #

When α=1, the simplified series becomes $\sum_{n=2}^\infty \frac{1}{n \ln n}$. Let's use the integral test here:
Take the function $f(x) = \frac{1}{x \ln x}$ for $x \geq 2$. It's positive, continuous, and steadily decreasing. Compute the improper integral:
$$\int_{2}^{\infty} \frac{1}{x \ln x} dx = \lim_{t \to \infty} \int_{2}^{t} \frac{1}{x \ln x} dx$$
Let $u = \ln x$, so $du = \frac{1}{x} dx$. The integral turns into:
$$\lim_{t \to \infty} \int_{\ln 2}^{\ln t} \frac{1}{u} du = \lim_{t \to \infty} \left[\ln u\right]{\ln 2}^{\ln t} = \lim{t \to \infty} (\ln(\ln t) - \ln(\ln 2)) = +\infty$$
Since the integral diverges, the series $\sum_{n=2}^\infty \frac{1}{n \ln n}$ diverges—so our original series does too.

Case 3: 0<α<1 #

Pick a number γ where α<γ<1. Let's take the limit of the ratio of our simplified term to $\frac{1}{n^\gamma}$:
$$\lim_{n \to \infty} \frac{\frac{1}{n^\alpha \ln n}}{\frac{1}{n^\gamma}} = \lim_{n \to \infty} \frac{n^{\gamma - \alpha}}{\ln n}$$
Since γ-α>0, $n^{\gamma - \alpha}$ grows way faster than $\ln n$, so this limit is $+\infty$. The series $\sum_{n=1}^\infty \frac{1}{n^\gamma}$ is a p-series with p=γ<1, which diverges. By the limit comparison test, if the ratio's limit is infinity and the comparison series diverges, our simplified series also diverges—so the original series does too.

Final Conclusion #

The infinite series $\sum_{n=1}^\infty \frac{1}{n^\alpha \ln(1+n^\beta)}$ converges if and only if α>1 (given β>1 and α>0).

内容的提问来源于stack exchange，提问作者Panda