Fundamentals of Infinite Series

Idea: use the definition of convergence and the relation $a_n = S_n - S_{n-1}$.

Details:

1. Assume $\sum_{n=1}^{\infty} a_n$ converges with sum $S$, i.e. $\lim_{n \to \infty} S_n = S,$ where $S_n = a_1 + a_2 + \cdots + a_n$.
2. Express $a_n$: for $n \ge 2$, $a_n = S_n - S_{n-1}$.
3. Take the limit: $\lim_{n \to \infty} a_n = \lim_{n \to \infty} (S_n - S_{n-1}) = S - S = 0.$
4. Conclusion: $\lim_{n \to \infty} a_n = 0$.

Note: This is necessary but not sufficient; $a_n \to 0$ does not guarantee convergence.

思路：Use the ratio test.

步骤：

1. Let $a_n = \frac{n}{2^n}$
2. Ratio: $\frac{a_{n+1}}{a_n} = \frac{n+1}{2^{n+1}} \cdot \frac{2^n}{n} = \frac{n+1}{2n}$
3. Limit: $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \frac{1}{2} < 1$
4. Conclude convergence.

答案：Convergent.

思路：Check absolute convergence.

步骤：

1. Consider $\sum_{n=1}^{\infty} \frac{1}{n^2}$
2. This is a $p$-series with $p = 2 > 1$, so it converges absolutely.
3. Therefore the original series converges.

答案：Convergent.

思路：Use the integral test.

步骤：

1. Let $f(x) = \frac{1}{x\ln x}$, so $a_n = f(n)$
2. Compute $\int_2^{+\infty} \frac{1}{x\ln x} dx = \ln(\ln x) \big|_2^{+\infty} = +\infty$
3. The integral diverges, so the series diverges.

答案：Divergent.

思路：Compare with a known convergent series.

步骤：

1. $\frac{1}{n^2 + n} < \frac{1}{n^2}$ and $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges ($p=2>1$)
2. By comparison, $\sum_{n=1}^{\infty} \frac{1}{n^2 + n}$ converges.

答案：Convergent.

思路：Test absolute convergence; if it fails, use Leibniz (alternating) test.

步骤：

1. Absolute series: $\sum_{n=1}^{\infty} \frac{1}{n^{1/2}}$ is a $p$-series with $p = \frac{1}{2} \le 1$, so diverges.
2. Original is alternating with $a_n = \frac{1}{n^{1/2}}$.
3. Check Leibniz conditions:
   • $a_n > 0$ ✓
   • $a_{n+1} < a_n$ ✓ (since $\frac{1}{(n+1)^{1/2}} < \frac{1}{n^{1/2}}$)
   • $\lim_{n \to \infty} a_n = 0$ ✓
4. Conditions hold, so the series converges conditionally.

答案：Convergent (conditional).