MOOCULUS微积分-2: 数列与级数学习笔记 3. Convergence tests

此课程(MOOCULUS-2 "Sequences and Series")由Ohio State University于2014年在Coursera平台讲授。

本系列学习笔记PDF下载（Academia.edu） MOOCULUS-2 Solution

Summary

Ratio Test
Consider the series $\sum_{n=0}^\infty a_n$ where each term $a_n$ is positive.
Suppose that $$\lim_{n\to \infty} \frac{a_{n+1}}{a_n}=L$$ Then,
- if $L < 1$ the series $\sum_{n=0}^\infty a_n$ converges.
- if $L > 1$ the series diverges.
- if $L=1$ this test is inconclusive.
Integral Test
Suppose that $f(x)>0$ and is decreasing on the infinite interval $[k,\infty)$ (for some $k\ge1$) and that $a_n=f(n)$. Then the series $$\sum_{n=k}^\infty a_n$$ converges if and only if the improper integral $$\int_{k}^\infty f(x)\,dx$$ converges.
$p$-series
Recall that a $p$-series is any series of the form $$\sum_{n=1}^\infty 1/n^p$$ A $p$-series converges if and only if $p>1$.
Root Test
Consider the series $\sum_{n=0}^\infty a_n$ where each term $a_n$ is positive. Suppose that $$\lim_{n\to \infty} \left(a_n\right)^{1/n}=L$$ Then,
- if $L < 1$ the series $\sum a_n$ converges,
- if $L > 1$ the series diverges, and
- if $L=1$, then the root test is inconclusive.
Determine a series convergent or divergent:
- $n^{\text{th}}$ test, if $$\lim_{n\to\infty}a_n\neq0$$ then it diverges; otherwise, using other tests;
- Comparison test, $p$-series test, ratio test, integral test, root test, and Cauchy test.

Exercises 3.1

1. Compute $$\lim_{n\to\infty} |a_{n+1}/a_n|$$ for the series $$\sum_{n=1}^\infty {1\over n^2}$$
Solution:
$$\lim_{n\to\infty} |a_{n+1}/a_n|=\lim_{n\to\infty}{{1\over(n+1)^2}\over{1\over n^2}}=\lim_{n\to\infty}{n^2\over n^2+2n+1}=1$$

2. Compute $$\lim_{n\to\infty} |a_{n+1}/a_n|$$ for the series $$\sum_{n=1}^\infty {1\over n}$$
Solution:
$$\lim_{n\to\infty} |a_{n+1}/a_n|=\lim_{n\to\infty}{{1\over n+1}\over{1\over n}}\lim_{n\to\infty}{n\over n+1}=1$$

Determine whether each of the following series converges or diverges.

3. $$\sum_{n=0}^\infty (-1)^{n}{3^n\over 5^n}$$
Solution:
By ratio test, we have $$\lim_{n\to\infty} |a_{n+1}/a_n|=\lim_{n\to\infty}{3^{n+1}\over 5^{n+1}}\cdot{5^n \over 3^n}={3\over5} < 1$$ Thus it converges.

4. $$\sum_{n=1}^\infty {n!\over n^n}$$
Solution:
By ratio test, we have
$$\lim_{n\to\infty} a_{n+1}/a_n=\lim_{n\to\infty}{(n+1)!\over(n+1)^{n+1}}\cdot{n^n\over n!}=\lim_{n\to\infty}{n^n\over(n+1)^n}=\lim_{n\to\infty}({n\over n+1})^n$$ On the other hand, $$\lim_{n\to\infty}(1+{1\over n})^n=e$$ Thus $$\lim_{n\to\infty}({n\over n+1})^n={1\over e} < 1$$ Therefore it converges.

5. $$\sum_{n=1}^\infty {n^5\over n^n}$$
Solution:
By ratio test, we have
$$\lim_{n\to\infty} a_{n+1}/a_n=\lim_{n\to\infty} {(n+1)^5\over(n+1)^{n+1}}\cdot{n^n\over n^5}$$ $$=\lim_{n\to\infty}({n+1 \over n})^5\cdot({n\over n+1})^n\cdot{1\over n+1}=1\times{1\over e}\times0=0 < 1$$ Thus it converges.

6. $$\sum_{n=1}^\infty {(n!)^2\over n^n}$$
Solution:
By ratio test, we have $$\lim_{n\to\infty} a_{n+1}/a_n=\lim_{n\to\infty}{((n+1)!)^2\over (n+1)^{n+1}}\cdot{n^n\over(n!)^2}$$ $$=\lim_{n\to\infty}{(n+1)^2\over(n+1)^{n+1}}\cdot n^n=\lim_{n\to\infty}(n+1)\cdot({n\over n+1})^n={1\over e}\cdot\lim_{n\to\infty}(n+1)\to\infty$$ Thus it diverges.

Exercises 3.2

Determine whether each series converges or diverges.

1. $$\sum_{n=1}^\infty {1\over n^{\pi/4}}$$
Solution:
$p$-series, and ${\pi\over4} < 1$ so it diverges.

2. $$\sum_{n=1}^\infty {n\over n^2+1}$$
Solution:
By integral test, we have
$$\int_{1}^\infty {n\over n^2+1}\,dn={1\over2}\int_{1}^{\infty}{d(n^2+1)\over n^2+1}={1\over2}\ln(n^2+1)\Big|_{1}^{\infty}\to\infty$$ Thus it diverges.

3. $$\sum_{n=1}^\infty {\ln n\over n^2}$$
Solution:
By integral test, we have $$u=\ln n,\ dv={1\over n^2}dn\Rightarrow du={1\over n}dn,\ v=-{1\over n}$$ $$\int_{1}^{\infty}{\ln n\over n^2}dn=-{\ln n\over n}\Big|_{1}^{\infty}+\int_{1}^{\infty}{1\over n^2}dn=-{\ln n\over n}\Big|_{1}^{\infty}-{1\over n}\Big|_{1}^{\infty}=0-(-1)=1$$ Thus it converges.

4. $$\sum_{n=1}^\infty {1\over n^2+1}$$
Solution:
$p$-series test, we have $$\sum_{n=1}^\infty {1\over n^2+1}<\sum_{n=1}^\infty {1\over n^2}\to\text{converges}$$ Thus it converges.

5. $$\sum_{n=1}^\infty {1\over e^n}$$
Solution:
This is geometric series and ${1\over e}< 1 $, thus it converges.

6. $$\sum_{n=1}^\infty {n\over e^n}$$
Solution:
By ratio test, we have
$$\lim_{n\to\infty}a_{n+1}/a_n=\lim_{n\to\infty}{n+1\over e^{n+1}}\cdot{e^n\over n}={1\over e}\lim_{n\to\infty}(1+{1\over n})={1\over e} < 1$$ Thus it converges.

7. $$\sum_{n=2}^\infty {1\over n\ln n}$$
Solution:
By integral test, we have
$$\int_{2}^{\infty}{1\over n\ln n} dn=\int_{2}^{\infty}{d(\ln n)\over\ln n}=\ln(\ln n)\Big|_{2}^{\infty}\to\infty$$ Thus it diverges.

8. $$\sum_{n=2}^\infty {1\over n(\ln n)^2}$$
Solution:
By integral test, we have
$$\int_{2}^{\infty}{1\over n(\ln n)^2}dn=\int_{2}^{\infty}{d(\ln n)\over(\ln n)^2}=-{1\over\ln n}\Big|_{2}^{\infty}={1\over\ln2}$$ Thus it converges.

9. Find an $N$ so that $$\sum_{n=1}^\infty {1\over n^4}$$ is between $$\sum_{n=1}^N {1\over n^4}$$ and $$\sum_{n=1}^N {1\over n^4} + 0.005$$
Solution:
For a series $\{a_n\}$, we define a continuous function $f(x)$ where $f(n)=a_n$. And we have $$R_n=\sum_{i=n+1}^{\infty}a_i=a_{n+1}+a_{n+2}+a_{n+3}+\cdots$$ Thus the series $$s=\sum_{i=1}^{\infty}a_1+a_2+\cdots+a_n+a_{n+1}+\cdots$$ $$=s_n+R_n=\sum_{i=1}^{n}a_i+\sum_{i=n+1}^{\infty}a_i$$ Particularly, $$\int_{n+1}^{\infty}f(x)dx < R_n < \int_{n}^{\infty}f(x)dx$$ Hence $$s_n+\int_{n+1}^{\infty}f(x)dx < s < s_n+\int_{n}^{\infty}f(x)dx$$ For this question, we have $$\int_{n+1}^{\infty}{1\over x^4}dx < R_n < 0.005$$ $$\Rightarrow -{1\over3}\cdot{1\over x^3}\Big|_{n+1}^{\infty}={1\over3}\cdot{1\over (n+1)^3} < 0.005$$ $$\Rightarrow n\geq4$$ R code:

## Q9

f = function(x) 1 / x^4

n = 1e7

limit = sum(f(x = 1:1e7))

for (i in 1:n){

  if(sum(f(x = 1:i)) < limit & limit < sum(f(x = 1:i)) + 0.005){

    print(i)

    break

  }

}

# [1] 4

10. Find an $N$ so that $$\sum_{n=0}^\infty {1\over e^n}$$ is between $$\sum_{n=0}^N {1\over e^n}$$ and $$\sum_{n=0}^N {1\over e^n} + 10^{-4}$$ Solution: $$\int_{n+1}^{\infty}{1\over e^x}dx < {1\over 10^4}\Rightarrow -e^{-x}\Big|_{n+1}^{\infty}=e^{-(n+1)} < 10^{-4}$$ $$\Rightarrow e^{n+1} > 10^4\Rightarrow n+1 > 4\ln10\Rightarrow n\geq9$$ R code:

## Q10

## Q10

f = function(x) 1 / exp(x)

n = 1e7

limit = sum(f(x = 0:1e7))

for (i in 0:n){

  if(sum(f(x = 0:i)) < limit & limit < sum(f(x = 0:i)) + 1e-4){

    print(i)

    break

  }

}

# [1] 9

11. Find an $N$ so that $$\sum_{n=1}^\infty {\ln n\over n^2}$$ is between $$\sum_{n=1}^N {\ln n\over n^2}$$ and $$\sum_{n=1}^N {\ln n\over n^2} + 0.005$$ Solution:
R code:

## Q11

f = function(x) log(x) / x^2

n = 1e7

limit = sum(f(x = 1:1e7))

for (i in 1:n){

  if(sum(f(x = 1:i)) < limit & limit < sum(f(x = 1:i)) + 0.005){

    print(i)

    break

  }

}

# [1] 1685

12. Find an $N$ so that $$\sum_{n=2}^\infty {1\over n(\ln n)^2}$$ is between $$\sum_{n=2}^N {1\over n(\ln n)^2}$$ and $$\sum_{n=2}^N {1\over n(\ln n)^2} + 0.005$$ Solution: $$\int_{n+1}^{\infty}{1\over x\cdot(\ln x)^2}dx < 0.005$$ $$\Rightarrow -{1\over\ln x}\Big|_{n+1}^{\infty}={1\over\ln(n+1)} < 0.005$$ $$\Rightarrow n+1 > e^{200}\Rightarrow n\geq e^{200}$$

Exercises 3.3

Determine whether the series converge or diverge.

1. $$\sum_{n=1}^\infty {1\over 2n^2+3n+5}$$ Solution:
When $n\geq1$ $${1\over 2n^2+3n+5} < {1\over 2n^2}\to\text{converge}$$ By $p$-series test and comparison test, it converges.

2. $$\sum_{n=2}^\infty {1\over 2n^2+3n-5}$$
Solution:
When $n\geq2$ $${1\over 2n^2+3n-5} < {1\over 2n^2}\to\text{converge}$$ By $p$-series test and comparison test, it converges.

3. $$\sum_{n=1}^\infty {1\over 2n^2-3n-5} $$
Solution:
When $n\geq5$ $${1\over 2n^2-3n-5}={1\over n^2+(n^2-3n-5)} < {1\over n^2}\to\text{converge}$$ By $p$-series test and comparison test, it converges.

4. $$\sum_{n=1}^\infty {3n+4\over 2n^2+3n+5} $$
Solution:
$$\int_{1}^{\infty}{3x+4\over 2x^2+3x+5}dx > \int_{1}^{\infty}{3x+{9\over4}\over 2x^2+3x+5}dx={3\over4}\cdot\int_{1}^{\infty}{d(2x^2+3x+5)\over 2x^2+3x+5}$$ $$={3\over4}\cdot\ln(2x^2+3x+5)\Big|_{1}^{\infty}\to\infty$$ By integral test and comparison test, it diverges.

5. $$\sum_{n=1}^\infty {3n^2+4\over 2n^2+3n+5} $$
Solution:
$$\lim_{n\to\infty}{3n^2+4\over 2n^2+3n+5}={3\over2}\neq0$$ By $n^{\text{th}}$ test, it diverges.

6. $$\sum_{n=1}^\infty {\log n\over n}$$
Solution:
When $n > e$ $${\log n\over n}>{1\over n}\to\text{diverge}$$ By $p$-series test and comparison test, it diverges.

7. $$\sum_{n=1}^\infty {\log n\over n^3}$$
Solution:
Set $u=\log x$, $dv={dx\over x^3}$, so $du={dx\over x}$, $v=-{1\over2}\cdot{1\over x^2}$. $$\int_{1}^{\infty}{\log x\over x^3}dx=-{1\over2}\cdot{1\over x^2}\cdot\log x\Big|_{1}^{\infty}+{1\over2}\int_{1}^{\infty}{dx\over x^3}$$ $$=-{1\over2}\cdot{1\over x^2}\cdot\log x\Big|_{1}^{\infty}-{1\over4}\cdot{1\over x^2}\Big|_{1}^{\infty}={1\over4}\to\text{converge}$$ By integral test, it converges.

8. $$\sum_{n=2}^\infty {1\over \log n}$$
Solution:
$$\log n < n\Rightarrow {1\over\log n} > {1\over n}\to\text{diverge}$$ By $p$-series test and comparison test, it diverges.

9. $$\sum_{n=1}^\infty {3^n\over 2^n+5^n}$$
Solution:
$$\lim_{n\to\infty}a_{n+1}/a_n=\lim_{n\to\infty}{3^{n+1}\over 2^{n+1}+5^{n+1}}\cdot{2^n+5^n \over 3^n}$$ $$=3\cdot\lim_{n\to\infty}{2^n+5^n \over 2^{n+1}+5^{n+1}}=3\cdot\lim_{n\to\infty}{({2\over5})^n+1 \over 2\cdot({2\over5})^n+5}={3\over5} < 1$$ By ratio test, it converges.

10. $$\sum_{n=1}^\infty {3^n\over 2^n+3^n}$$
Solution:
$$\lim_{n\to\infty}{3^n\over 2^n+3^n}=\lim_{n\to\infty}{1\over ({2\over3})^n+1}=1\neq0$$ By $n^{\text{th}}$ test, it diverges.

Exercises 3.4

1. Prove the root test theorem.
Solution:
First, prove when $L < 1$ it converges. There should be an $r$ such that $L < r < 1$, and there exists some $n$, when $n < N$ we have $${a_n}^{{1\over n}}=L < r < 1\Rightarrow a_n < r^n\to\text{converge}$$ Note that the last expression is a geometric series. By comparison test, $\{a_n\}$ converges.
Secondly we prove when $L > 1$ it diverges.
There exists some $n > N$ we have $${a_n}^{{1\over n}}=L > 1\Rightarrow a_n > 1^n=1$$ which means $$\lim_{n\to\infty}a_n > 1\neq0$$ By $n^{\text{th}}$ test, it diverges.
Lastly, when $L=1$ it is inconclusive. Such as

$\lim_{n\to\infty}({1\over n})^{{1\over n}}=1$, and it diverges.
$\lim_{n\to\infty}({1\over n^2})^{{1\over n}}=1$, and it converges.

2. Compute $$\lim_{n\to\infty} |a_n|^{1/n}$$ for the series $\sum 1/n^2$.
Solution:
$$\lim_{n\to\infty} ({1\over n^2})^{1/n}=1$$ But it converges by $p$-series test.

3. Compute $$\lim_{n\to\infty} |a_n|^{1/n}$$ for the series $\sum 1/n$.
Solution:
$$\lim_{n\to\infty} ({1\over n})^{1/n}=1$$ By $p$-series test, it converges.

Additional Exercises

Determine whether the series converge or diverge.

1. $$\sum_{n=6}^{\infty}{15^n \over 3^{3n+3}\cdot(3n+4)}$$
Solution:
$$\lim_{n\to\infty}a_{n+1}/a_n=\lim_{n\to\infty}{15^{n+1}\over 3^{3n+6}\cdot(3n+7)}\cdot{3^{3n+3}\cdot(3n+4)\over 15^n}$$ $$=\lim_{n\to\infty}{15\cdot(3n+4)\over 27\cdot(3n+7)}={5\over9} < 1$$ By ratio test, it converges.

2. $$\sum_{k=8}^{\infty}{(k+2)\cdot k!\over 4^k}$$
Solution:
$$\lim_{k\to\infty}a_{n+1}/a_n=\lim_{k\to\infty}{(k+3)\cdot(k+1)!\over4^{k+1}}\cdot{4^{k}\over(k+2)\cdot k!}=\lim_{k\to\infty}{(k+3)(k+1)\over4(k+2)}\to\infty$$ By ratio test, it diverges.

3. $$\sum_{n=1}^{\infty}{10^n\cdot n!\over(9n)^n}$$
Solution:
$$\lim_{n\to\infty}a_{n+1}/a_n=\lim_{n\to\infty}{10^{n+1}\cdot(n+1)!\over9^{n+1}\cdot(n+1)^{n+1}}\cdot{9^n\cdot n^n\over10^n\cdot n!}$$ $$=\lim_{n\to\infty}{10(n+1)\cdot n^n\over9(n+1)^{n+1}}={10\over9}\cdot({n\over n+1})^n={10\over9}\cdot{1\over e} < 1$$ By ratio test, it converges.

4. $$\sum_{k=3}^{\infty}{3\over7k^{1.88}}$$
Solution:
$$\sum_{k=3}^{\infty}{3\over7k^{1.88}}=\sum_{k=3}^{\infty}{3\over7}\cdot{1\over k^{1.88}}$$ By $p$-series test, it converges.

5. $$\sum_{i=6}^{\infty}{i-2\over i^5+3i}$$
Solution:
$${i-2\over i^5+3i} < {i\over i^5+3i} < {i\over i^5}={1\over i^4}$$ By $p$-series test and comparison test, it converges.

6. $$\sum_{m=2}^{\infty}({7m+5 \over 6m+5})^m$$
Solution:
$$\lim_{m\to\infty}{a_m}^{{1\over m}}=\lim_{m\to\infty}{7m+5\over 6m+5}={7\over6} > 1$$ By root test, it diverges.

7. $$\int_{2}^{\infty}{4\over (4x+1)^2}dx={1\over9}$$
What is the interval of $$\sum_{k=2}^{\infty}{4\over(4k+1)^2}$$
Solution:
Since $$\int_{m}^{\infty}f(x)dx \leq \sum_{n=m}^{\infty}a_n \leq a_m+\int_{m}^{\infty}f(x)dx$$ That is $$\int_{2}^{\infty}{4\over(4x+1)^2}dx\leq\sum_{k=2}^{\infty}{4\over(4k+1)^2}\leq a_2+\int_{2}^{\infty}{4\over(4x+1)^2}dx$$ $$\Rightarrow {1\over9}\leq s_k\leq{4\over81}+{1\over9}={13\over81}$$ Thus the interval is $[{1\over9},\ {13\over81}]$.

巴特西

MOOCULUS微积分-2: 数列与级数学习笔记 3. Convergence tests

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