PG5.5 Stage review: proving a series converges

Simulate

Check the first claim with numbers

This review asks for three proofs, each with a different tool from the stage. The first series is $\sum_{k=1}^{\infty} \frac{k^2}{2^k}$ : the tops grow, but the bottoms grow faster.

The cell prints the ratio $\frac{a_{k+1}}{a_k} = \frac{(k+1)^2}{2k^2}$ at a few values of $k$ , then partial sums.

for k in [1, 2, 5, 10, 100, 1000]: print(k, round((k + 1)**2 / (2 * k**2), 4)) for n in [10, 20, 40, 60]: print(n, round(sum(k**2 / 2**k for k in range(1, n + 1)), 8))

⌘/Ctrl + Enter runs the cell

The ratios fall from 2.0 towards 0.5: 0.605 at $k = 10$ , 0.501 at $k = 1000$ . The partial sums settle on 6.0, to eight decimal places from 40 terms on.

Conjecture

\sum_{k=1}^{\infty} \frac{k^2}{2^k}

Six ratios and four partial sums. The proof needs the limit of the ratio, worked out for every $k$ .

The tools of this stage

Each proof below needs one or two of these. Name the one you use, and check that its conditions hold before you apply it.

tool	what it says	lesson
term test	$\sum a_k$ converges $\Rightarrow$ $a_k \to 0$	PG5.1
nonnegative terms	bounded partial sums $\Rightarrow$ converges	PG5.2
comparison	$0 \le a_k \le b_k$ , $\sum b_k$ converges $\Rightarrow$ $\sum a_k$ converges	PG5.2
ratio	$a_k > 0$ , $\frac{a_{k+1}}{a_k} \to L < 1$ $\Rightarrow$ converges	PG5.2
absolute convergence	$\sum \lvert a_k \rvert$ converges $\Rightarrow$ $\sum a_k$ converges	PG5.3
fast factorials	$\frac{\lvert x \rvert^n}{n!} \to 0$ for every $x$	PG5.4

Format 4

Write it

Prove that $\sum_{k=1}^{\infty} \frac{k^2}{2^k}$ converges. Work out the ratio $\frac{a_{k+1}}{a_k}$ , find its limit and say why, then apply a test.

Your proof

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Write your own first: a proof you have tried to write teaches far more than one you only read.

Simulate

Check the second claim

The second series has signs that follow no pattern: $\sum_{k=1}^{\infty} \frac{\cos k}{k^2}$ , with $k$ in radians.

import numpy as np k = np.arange(1, 10**6 + 1, dtype=float) S = np.cumsum(np.cos(k) / k**2) for n in (10, 100, 1000, 10**6): print(n, round(float(S[n - 1]), 6)) print(bool(np.all(np.abs(np.cos(k)) / k**2 <= 1 / k**2))) print(round(float(np.sum(np.abs(np.cos(k)) / k**2)), 6))

⌘/Ctrl + Enter runs the cell

The partial sums settle near 0.324138. Every size $\frac{|\cos k|}{k^2}$ is at most $\frac{1}{k^2}$ up to a million, and the sizes add up to 0.927641 over those terms.

Conjecture

\sum_{k=1}^{\infty} \frac{\cos k}{k^2}

A million terms. The ratio test cannot help: the terms change sign, and the ratios of their sizes jump about. The proof needs another tool.

Format 4

Write it

Prove that $\sum_{k=1}^{\infty} \frac{\cos k}{k^2}$ converges, and that its sum lies between $-2$ and 2.

Your proof

Saved in this browser as you type.

Write your own first: a proof you have tried to write teaches far more than one you only read.

Simulate

Check the third claim

Lesson 2.7.2 gave the series $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$ for every $x$ , in radians. Write $p_n(x)$ for its Taylor polynomial of degree $n$ . The cell prints $|\sin x - p_n(x)|$ next to $\frac{|x|^{n+1}}{(n+1)!}$ .

import math from math import factorial def p_sin(x, n): return sum((-1)**j * x**(2 * j + 1) / factorial(2 * j + 1) for j in range(n // 2 + 1) if 2 * j + 1 <= n) for x, n in [(3, 5), (3, 11), (3, 21), (10, 11), (10, 21), (10, 31)]: error = abs(math.sin(x) - p_sin(x, n)) bound = abs(x)**(n + 1) / factorial(n + 1) print(x, n, f"{error:.3e}", f"{bound:.3e}")

⌘/Ctrl + Enter runs the cell

Every error is below $\frac{|x|^{n+1}}{(n+1)!}$ . At $x = 10$ the polynomial of degree 11 is still off by $1.056 \times 10^{3}$ , but by degree 31 the error is $1.062 \times 10^{-4}$ .

Conjecture

x

Two values of $x$ . The proof follows lesson PG5.4, with a better $M$ .

Format 4

Write it

Prove that $p_n(x) \to \sin x$ for every real $x$ , where $p_n$ is the Taylor polynomial of $\sin x$ of degree $n$ about 0. Use the remainder bound of lesson 2.7.2, and say what $M$ is.

Your proof

Saved in this browser as you type.

Write your own first: a proof you have tried to write teaches far more than one you only read.

The cosine series works the same way, with $M = 1$ again. Stage PG6 turns from series to sets, maps and sizes.

\frac{(k+1)^2}{2k^2} = \frac{1}{2}\left(1 + \frac{1}{k}\right)^2 \to \frac{1}{2}

Stage review: proving a series converges

Check the first claim with numbers

The tools of this stage

Write it

Check the second claim

Write it

Check the third claim

Write it

Summary card

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