Determine if the series converges:判断该级数是否收敛：
$$ \sum_{n=1}^{\infty} \frac{1}{n(n+1)} $$

Use partial fractions:使用部分分式分解：
  1/[n(n+1)] = 1/n − 1/(n+1)

Write partial sums (terms cancel!):写出部分和（各项相消！）：
  S_n = (1 − 1/2) + (1/2 − 1/3) + ⋯ + (1/n − 1/(n+1))
      = 1 − 1/(n+1)

Take the limit:取极限：
  lim(n→∞) S_n = 1 − 0 = 1

The series converges to 1.该级数收敛到 1。

Find the sum:求和：
$$ \sum_{n=0}^{\infty} \frac{(-1)^n \cdot 2^n}{3^n} = \sum_{n=0}^{\infty} \left(\frac{-2}{3}\right)^n $$

Identify: a = 1, r = −2/3识别：a = 1，r = −2/3
Since |r| = 2/3 < 1, it converges:因为 |r| = 2/3 < 1，所以收敛：

  S = a/(1−r) = 1/(1−(−2/3))
    = 1/(1 + 2/3) = 1/(5/3) = 3/5

Does this series converge or diverge?该级数收敛还是发散？
$$ \sum_{n=1}^{\infty} \frac{n}{2n+1} $$

Apply nth term test:应用第 n 项判别法：
  lim(n→∞) n/(2n+1) = lim(n→∞) 1/(2 + 1/n) = 1/2 ≠ 0

By the nth Term Test, the series diverges.由第 n 项判别法，级数发散。

Does the series converge?该级数是否收敛？
$$ \sum_{n=1}^{\infty} \frac{1}{n^2 + 1} $$

Let f(x) = 1/(x² + 1). It is positive, continuous,设 f(x) = 1/(x² + 1)。它在 x ≥ 1 上
and decreasing for x ≥ 1. ✓为正、连续且单调递减。✓

Evaluate the improper integral:计算反常积分：
∫₁∞ 1/(x²+1) dx = lim(b→∞) [arctan(x)]₁ᵇ
                 = lim(b→∞) [arctan(b) − arctan(1)]
                 = π/2 − π/4 = π/4

The integral converges, so the series converges.积分收敛，因此级数收敛。
(Note: the series sum ≠ π/4; the integral test（注意：级数和 ≠ π/4；积分判别法
 only tells us convergence, not the exact sum.) 只判断收敛性，不给出精确和。）

Does this converge?该级数是否收敛？
$$ \sum_{n=1}^{\infty} \frac{3n+1}{n^3 - 2} $$

For large n, the dominant behavior is ≈ 3n/n³ = 3/n².当 n 很大时，主导项 ≈ 3n/n³ = 3/n²。
Compare with b_n = 1/n² (p-series, p=2, converges).与 b_n = 1/n² 比较（p 级数，p=2，收敛）。

  lim(n→∞) a_n/b_n = lim(n→∞) [(3n+1)/(n³−2)] / [1/n²]
                    = lim(n→∞) (3n+1)·n²/(n³−2)
                    = lim(n→∞) (3n³+n²)/(n³−2)
                    = 3

Since 0 < 3 < ∞ and Σ 1/n² converges:因为 0 < 3 < ∞ 且 Σ 1/n² 收敛：
By the Limit Comparison Test, the series converges.由极限比较判别法，级数收敛。

Does this converge?该级数是否收敛？
$$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2} $$

Here b_n = 1/n².此处 b_n = 1/n²。

✓ b_n > 0 for all n✓ 对所有 n，b_n > 0
✓ b_{n+1} = 1/(n+1)² < 1/n² = b_n (decreasing)✓ b_{n+1} = 1/(n+1)² < 1/n² = b_n（递减）
✓ lim(n→∞) 1/n² = 0✓ lim(n→∞) 1/n² = 0

All three conditions met → series converges三个条件全部满足 → 由交错级数判别法，
by the Alternating Series Test.级数收敛。

Does this converge?该级数是否收敛？
$$ \sum_{n=1}^{\infty} \frac{n!}{3^n} $$

Compute the ratio:计算比值：
  |a_{n+1}/a_n| = [(n+1)!/3^(n+1)] / [n!/3^n]
               = (n+1)!/n! · 3^n/3^(n+1)
               = (n+1) · 1/3
               = (n+1)/3

  L = lim(n→∞) (n+1)/3 = ∞ > 1

By the Ratio Test, the series diverges.由比值判别法，级数发散。
(Factorials grow faster than exponentials!)（阶乘增长比指数更快！）

Does this converge?该级数是否收敛？
$$ \sum_{n=0}^{\infty} \frac{5^n}{n!} $$

Compute the ratio:计算比值：
  |a_{n+1}/a_n| = [5^(n+1)/(n+1)!] / [5^n/n!]
               = 5/(n+1)

  L = lim(n→∞) 5/(n+1) = 0 < 1

By the Ratio Test, the series converges absolutely.由比值判别法，级数绝对收敛。
(This is actually the series for e⁵!)（这其实就是 e⁵ 的级数！）

Approximate the sum of the series to within 0.01:将该级数之和近似到误差 0.01 以内：
$$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^3} $$

We need |a_{n+1}| = 1/(n+1)³ < 0.01需要 |a_{n+1}| = 1/(n+1)³ < 0.01
  (n+1)³ > 100
  n+1 > 4.64...
  n ≥ 4  →  use 使用 S₄

Compute:计算：
  S₄ = 1/1 − 1/8 + 1/27 − 1/64
     = 1 − 0.125 + 0.0370 − 0.0156
     ≈ 0.8965

S₄ approximates the sum with error ≤ 1/125 = 0.008 < 0.01. ✓S₄ 近似该和，误差 ≤ 1/125 = 0.008 < 0.01。✓

Find the 4th-degree Maclaurin polynomial for f(x) = eˣ.求 f(x) = eˣ 的 4 次麦克劳林多项式。

All derivatives of eˣ are eˣ, and e⁰ = 1:eˣ 的各阶导数都是 eˣ，且 e⁰ = 1：
  f(0) = 1,  f'(0) = 1,  f''(0) = 1,  f'''(0) = 1,  f⁴(0) = 1

P₄(x) = 1 + x + x²/2! + x³/3! + x⁴/4!
       = 1 + x + x²/2 + x³/6 + x⁴/24

Use the 3rd-degree Maclaurin polynomial for sin(x)用 sin(x) 的 3 次麦克劳林多项式
to approximate sin(0.5). Bound the error.近似 sin(0.5)，并估计误差界。

P₃(x) = x − x³/6
P₃(0.5) = 0.5 − 0.125/6 = 0.5 − 0.02083 ≈ 0.47917

Error bound: n = 3, a = 0, x = 0.5误差界：n = 3，a = 0，x = 0.5
f⁴(x) = sin(x), so |f⁴(c)| ≤ 1 for all c → M = 1f⁴(x) = sin(x)，所以对所有 c 有 |f⁴(c)| ≤ 1 → M = 1

  |R₃| ≤ M · |x−a|⁴ / 4! = 1 · (0.5)⁴ / 24
       = 0.0625/24 ≈ 0.0026

The approximation 0.47917 has error ≤ 0.0026.近似值 0.47917 的误差 ≤ 0.0026。
(Actual: sin(0.5) ≈ 0.47943, error ≈ 0.00026)（实际：sin(0.5) ≈ 0.47943，误差 ≈ 0.00026）

Find the interval of convergence:求收敛区间：
$$ \sum_{n=1}^{\infty} \frac{(x-3)^n}{n \cdot 2^n} $$

Ratio test:比值判别法：
  |a_{n+1}/a_n| = |(x−3)^(n+1)/[(n+1)2^(n+1)]| / |(x−3)^n/[n·2^n]|
               = |x−3|/2 · n/(n+1)

  lim = |x−3|/2

Converges when |x−3|/2 < 1  →  |x−3| < 2  →  R = 2当 |x−3|/2 < 1 时收敛  →  |x−3| < 2  →  R = 2
Open interval: (1, 5)开区间：(1, 5)

Test x = 1:检验 x = 1：  Σ (-1)^n/n → alternating harmonic → 交错调和级数 → converges收敛
Test x = 5:检验 x = 5：  Σ 1/n → harmonic → 调和级数 → diverges发散

Interval of convergence: [1, 5)收敛区间：[1, 5)

Find the Maclaurin series for f(x) = e^(−x²).求 f(x) = e^(−x²) 的麦克劳林级数。

Start with the known series for eᵘ:从已知的 eᵘ 级数出发：
  eᵘ = Σ uⁿ/n!

Substitute u = −x²:代入 u = −x²：
  e^(−x²) = Σ (−x²)ⁿ/n!
           = Σ (−1)ⁿ x^(2n) / n!

           = 1 − x² + x⁴/2! − x⁶/3! + ⋯

           = Σ_{n=0}^∞ (−1)ⁿ x^(2n) / n!

Find a power series for 1/(1+x²).求 1/(1+x²) 的幂级数。

Start with: 1/(1−u) = Σ uⁿ for |u| < 1从已知出发：当 |u| < 1 时 1/(1−u) = Σ uⁿ
Let u = −x²:令 u = −x²：

  1/(1−(−x²)) = 1/(1+x²)
              = Σ (−x²)ⁿ
              = Σ (−1)ⁿ x^(2n)

              = 1 − x² + x⁴ − x⁶ + ⋯
for |x| < 1|x| < 1

Find a power series for arctan(x).求 arctan(x) 的幂级数。

We know d/dx[arctan(x)] = 1/(1+x²)已知 d/dx[arctan(x)] = 1/(1+x²)
So arctan(x) = ∫ 1/(1+x²) dx所以 arctan(x) = ∫ 1/(1+x²) dx

From above: 1/(1+x²) = Σ (−1)ⁿ x^(2n)由上：1/(1+x²) = Σ (−1)ⁿ x^(2n)

Integrate term by term:逐项积分：
  arctan(x) = Σ (−1)ⁿ x^(2n+1)/(2n+1) + C

Since arctan(0) = 0, we get C = 0:因为 arctan(0) = 0，所以 C = 0：
  arctan(x) = x − x³/3 + x⁵/5 − x⁷/7 + ⋯
for |x| ≤ 1|x| ≤ 1