↑ Year 1 · Semester 1

MH1805

Calculus

Series

Definition

With a sequence {ak}k=1∞, we can form the nth partial sum Sn=∑k=1nak=a1+a2+⋯+an.\text{With a sequence } \{a_k\}_{k=1}^{\infty},\text{ we can form the } n\text{th partial sum }S_n = \sum_{k=1}^{n} a_k = a_1 + a_2 + \cdots + a_n. If limn→∞Sn exists, the series ∑k=1∞ak converges.\text{If } \lim_{n \to \infty} S_n \text{ exists, the series } \sum_{k=1}^{\infty} a_k \text{ converges.}

*Series is the sum of the terms of the sequence.

Sum Law for Series

∑k=1∞(Aak+Bbk)=A∑k=1∞ak+B∑k=1∞bk, if both series converge.\sum_{k=1}^{\infty} (A a_k + B b_k)= A \sum_{k=1}^{\infty} a_k+ B \sum_{k=1}^{\infty} b_k \text{, if both series converge.}

If one diverges and one converges, the sum of them diverges.

If both diverge, no conclusion.

Series of positive terms

If {ak}k=1∞ is an increasing sequence, then limk→∞ak exists ⇔{ak} is bounded.\text{If } \{a_k\}_{k=1}^{\infty} \text{ is an increasing sequence, then }\lim_{k \to \infty} a_k \text{ exists } \Longleftrightarrow \{a_k\} \text{ is bounded.}

If f:[a,∞)→ℝ is continuous and increasing, then limx→+∞f(x) exists ⇔f is bounded.\text{If } f:[a,\infty) \to \mathbb{R} \text{ is continuous and increasing, then }\lim_{x \to +\infty} f(x) \text{ exists } \Longleftrightarrow f \text{ is bounded.}

For Sn=∑k=1∞ak, and ak≥0 for k=1,2,3,…, then\text{For } S_n = \sum_{k=1}^{\infty} a_k, \text{ and } a_k \ge 0 \text{ for } k=1,2,3,\ldots, \text{ then}

∑k=1∞ak converges ⇔(Sn)n=1∞ is a bounded sequence.\sum_{k=1}^{\infty} a_k \text{ converges } \Leftrightarrow \ (S_n)_{n=1}^{\infty} \text{ is a bounded sequence.}

Tests

Limit Test

If the series ∑n=1∞an converges, then limn→∞an=0.\text{If the series } \sum_{n=1}^{\infty} a_n \text{ converges, then } \lim_{n \to \infty} a_n = 0.

Contrapositive: If an↛0, then the series ∑n=1∞an does not converge.\text{Contrapositive: If } a_n \not\to 0, \text{ then the series } \sum_{n=1}^{\infty} a_n \text{ does not converge.}

Integral test

If f:[1,∞)→ℝ is continuous, decreasing, and limx→+∞f(x)=0, then\text{If } f:[1,\infty)\to\mathbb{R} \text{ is continuous, decreasing, and } \lim_{x\to +\infty} f(x)=0,\text{ then}

∑k=1∞f(k) converges ⇔∫1+∞f(x)dx converges.\sum_{k=1}^{\infty} f(k) \text{ converges } \Leftrightarrow \ \int_{1}^{+\infty} f(x)\,dx \text{ converges.}

P-series

∑n=1∞1np converges ⇔p>1.\sum_{n=1}^{\infty} \frac{1}{n^p} \text{ converges } \Leftrightarrow \ p > 1.

Harmonic series ∑n=1∞1n diverges.\text{Harmonic series } \sum_{n=1}^{\infty} \frac{1}{n} \text{ diverges.}

For series with positive terms

General comparison test

If 0≤an≤Cbn, for some constant C>0 and any positive integers n, then\text{If } 0 \le a_n \le C b_n, \text{ for some constant } C > 0 \text{ and any positive integers } n,\text{ then}

if ∑bn converges, then ∑an must also converge.\text{if } \sum b_n \text{ converges, then } \sum a_n \text{ must also converge.}

if ∑an diverges, then ∑bn must also diverge.\text{if } \sum a_n \text{ diverges, then } \sum b_n \text{ must also diverge.}

**terms must be non-negative

Limit Comparison Test

For an≥0,bn>0 for all n,if limn→∞anbn=L,where L>0,L∈ℝ,\text{For } a_n \ge 0,\, b_n > 0 \text{ for all } n,\ \text{if } \lim_{n \to \infty} \frac{a_n}{b_n} = L,\ \text{where } L > 0,\ L \in \mathbb{R},

then ∑n=1∞an and ∑n=1∞bn either both converge or both diverge.\text{then } \sum_{n=1}^{\infty} a_n \text{ and } \sum_{n=1}^{\infty} b_n \text{ either both converge or both diverge.}

Ratio Test

For an>0 for large n, and limn→∞|an+1an|=c,\text{For } a_n > 0 \text{ for large } n, \text{ and } \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = c,

if c>1, then ∑an diverges.\text{if } c > 1, \text{ then } \sum a_n \text{ diverges.}

if c<1, then ∑an converges.\text{if } c < 1, \text{ then } \sum a_n \text{ converges.}

if c=1, gives no information.\text{if } c = 1, \text{ gives no information.}

Root Test

For an≥0 for large n, and limn→∞ann=c,\text{For } a_n \ge 0 \text{ for large } n, \text{ and } \lim_{n \to \infty} \sqrt[n]{a_n} = c,

if c>1, then ∑an diverges.\text{if } c > 1, \text{ then } \sum a_n \text{ diverges.}

if c<1, then ∑an converges.\text{if } c < 1, \text{ then } \sum a_n \text{ converges.}

if c=1, no information can be concluded.\text{if } c = 1, \text{ no information can be concluded.}

For Series with both positive and negative terms

Absolute convergence

If ∑n=1∞|an| converges, then ∑n=1∞an converges absolutely.\text{If } \sum_{n=1}^{\infty} |a_n| \text{ converges, then } \sum_{n=1}^{\infty} a_n \text{ converges absolutely.}

If ∑n=1∞an converges absolutely, then it converges.\text{If } \sum_{n=1}^{\infty} a_n \text{ converges absolutely, then it converges.}

**If the series converges absolutely, then any series obtained from it by rearranging its elements also converges to the same sum.

Conditional convergence

If ∑n=1∞an converges, but ∑n=1∞|an| does not, then ∑n=1∞an converges conditionally.\text{If } \sum_{n=1}^{\infty} a_n \text{ converges, but } \sum_{n=1}^{\infty} |a_n| \text{ does not, then } \sum_{n=1}^{\infty} a_n \text{ converges conditionally.}

Riemann’s Theorem

If a series converges conditionally, then we can rearrange its elements to make the sum any other number or even diverge.

Reason: The individual terms are small (the series converges), but the positive terms diverge and the negative terms diverge. Given any sum S, we can add positive terms to just exceed S, then subtract negative terms to just dip below S. Repeat forever and the sum settles to S as the terms get smaller.

Example (alternating harmonic series):

∑n=1∞(−1)n−1n=1−12+13−14+15−⋯=ln⁡2\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \cdots = \ln 2

Consider the rearrangement:

Sum=1−12−14+13−16−18+15−110−112+⋯\text{Sum} = 1 - \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{6} - \frac{1}{8} + \frac{1}{5} - \frac{1}{10} - \frac{1}{12} + \cdots

=(1−12)−14+(13−16)−18+(15−110)−112+⋯= \left(1 - \frac{1}{2}\right) - \frac{1}{4} + \left(\frac{1}{3} - \frac{1}{6}\right) - \frac{1}{8} + \left(\frac{1}{5} - \frac{1}{10}\right) - \frac{1}{12} + \cdots

=12−14+16−18+110−112+⋯= \frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8} + \frac{1}{10} - \frac{1}{12} + \cdots

=12(1−12+13−14+15−16+⋯)= \frac{1}{2} \left( 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots \right)

=12ln⁡2= \frac{1}{2} \ln 2

Leibniz’ Test for alternating series

For an alternating series ±∑n=1∞(−1)n−1bn, where bn>0,\text{For an alternating series } \pm \sum_{n=1}^{\infty} (-1)^{n-1} b_n, \text{ where } b_n > 0,

if b1≥b2≥b3≥…≥0 and limn→∞bn=0,\text{if } b_1 \ge b_2 \ge b_3 \ge \ldots \ge 0 \text{ and } \lim_{n \to \infty} b_n = 0,

then ∑n=1∞(−1)n−1bn converges.\text{then } \sum_{n=1}^{\infty} (-1)^{n-1} b_n \text{ converges.}

Alternating harmonic series ∑n=1∞(−1)n−1n converges conditionally\text{Alternating harmonic series }\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \text{ converges conditionally}