Solution Manual of MOOCULUS-2 "Sequences and Series": Review and Final

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Review

Determine whether the series converges.

1. $\displaystyle\sum_{n=0}^{\infty}{n\over n^2+4}$
Solution:
When $n > 4$ we have
$${n\over n^2+4} > {n\over n^2+n} = {1\over n+1}\to\text{diverge}$$ By $p$-series test and comparison test, it diverges.

2. $\displaystyle{1\over1\cdot2}+{1\over3\cdot4}+{1\over5\cdot6} +{1\over7\cdot8}+\cdots\cdots$
Solution:
$$s=\sum_{n=0}^{\infty}{1\over (2n+1)(2n+2)}$$ and $${1\over (2n+1)(2n+2)} = {1\over 4n^2+6n+2} < {1\over n^2}\to\text{converge}$$ By $p$-series test and comparison test, it converges.

3. $\displaystyle\sum_{n=0}^{\infty}{n\over(n^2+4)^2}$  
Solution:
$${n\over(n^2+4)^2} < {n\over (n^2)^2} = {1\over n^3}\to\text{converge}$$ By $p$-series test and comparison test, it converges.

4. $\displaystyle\sum_{n=0}^{\infty}{n!\over8^n}$  
Solution:
$$\lim_{n\to\infty}a_{n+1}/a_n={(n+1)!\over8^{n+1}}\cdot{8^n\over n!}={n+1\over8}\to\infty$$ By ratio test, it diverges.

5. $\displaystyle1-{3\over4}+{5\over8}-{7\over12}+{9\over16}+ \cdots\cdots$
Solution:
$$s=1+\sum_{n=1}^{\infty}(-1)^n{2n+1\over4n}$$ and $$\lim_{n\to\infty}{2n+1\over4n}={1\over2}\neq0$$ Thus it diverges.

Solution Manual of MOOCULUS-2 "Sequences and Series": 7. Taylor series

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Summary

• Given a function $f$, the series $$\sum_{n=0}^\infty {f^{(n)}(0)\over n!}x^n$$ is called the Maclaurin series for $f$, or often just the Taylor series for $f$ centered around zero.
• Given a function $f$, the series $$\sum_{n=0}^\infty {f^{(n)}(c)\over n!}(x-c)^n$$ is called the Taylor series for $f$ centered around $c$.
• Taylor's Theorem
  Suppose that $f$ is defined on some open interval $I = (a-R,a+R)$ around $a$ and suppose the function $f$ is $(N+1)$-times differentiable on $I$, meaning that $f^{(N+1)}(x)$ exists for $x\in I$. Then for each $x \neq a$ in $I$ there is a value $z$ between $x$ and $a$ so that $$f(x) = \sum_{n=0}^N {f^{(n)}(a)\over n!}\,(x-a)^n + {f^{(N+1)}(z)\over (N+1)!}(x-a)^{N+1}.$$
• Common functions:
  $$e^x=\sum_{n=0}^{\infty}{1\over n!}x^n=1+x+{x^2\over2!}+{x^3\over3!}+\cdots\cdots,\ \text{for all}\ x$$ $${1\over1-x}=\sum_{n=0}^{\infty}x^n=1+x+x^2+x^3+\cdots\cdots,\ \ \text{for}\ |x| < 1$$ $$\log(1+x)=\sum_{n=1}^{\infty}{(-1)^{n-1}\over n}x^n=x-{1\over2}x^2+{1\over3}x^3+\cdots\cdots,\ \ \text{for}\ -1 < x \leq 1$$ $$\sin x=\sum_{n=0}^{\infty}{(-1)^n\over(2n+1)!}x^{2n+1}=x-{x^3\over3!}+{x^5\over5!} +\cdots\cdots,\ \text{for all}\ x$$ $$\cos x=\sum_{n=0}^{\infty}{(-1)^n\over(2n)!}x^{2n}=1-{x^2\over2!}+{x^4\over4!} +\cdots\cdots,\ \text{for all}\ x$$

Solution Manual of MOOCULUS-2 "Sequences and Series": 6. Power series

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Summary

• Let $(a_n)$ be a sequence of real numbers starting with $a_0$. Then the power series associated to $(a_n)$ is $$\sum_{n=0}^\infty a_n \, x^n.$$
  Note that $a_n$ does not depend on $x$.
• The set of values of $x$ for which the series $$\sum_{n=0}^\infty a_n \, x^n$$ converges is the interval of convergence.
  That is, by ratio test we have $$\lim_{n\to\infty}{|a_{n+1}\cdot x^{n+1}|\over|a_n\cdot x^{n}|}=|x|\cdot\lim_{n\to\infty}{|a_{n+1}|\over|a_n|} < 1$$ it will converge. Technically, $${1\over R}=\lim_{n\to\infty}{|a_{n+1}|\over |a_n|}$$
• For a power series, the interval of convergence is, in fact, an interval. It has the form $(-R,R)$ or $[-R,R)$ or $(-R,R]$ or $[-R,R]$. In short, it is centered around $0$.
• In the interval of convergence of a power series, the value $R$ is called the radius of convergence of the series.
• Let $(a_n)$ be a sequence of real numbers starting with $a_0$. Then the power series centered at $c$ and associated to $(a_n)$ is the series
  $$\sum_{n=0}^\infty a_n \, (x-c)^n.$$ That is, the interval of convergence is $I=(c-R, c+R)$ (or include the endpoints).
• Suppose the power series $$f(x)=\sum_{n=0}^\infty a_n(x-a)^n=a_0+a_1\cdot(x-a)+a_2\cdot(x-a)^2+\cdots$$ has radius of convergence $R$. Then $$f'(x)=a_1+2a_2\cdot(x-a)+\cdots=\sum_{n=1}^\infty na_n(x-a)^{n-1}$$ $$\int f(x)\,dx = C+\sum_{n=0}^\infty {a_n\over n+1}(x-a)^{n+1}$$
  for interval of $x$ in the interval $(a-R, a+R)$. These two new series have radius of convergence $R$, just like the original series.

Solution Manual of MOOCULUS-2 "Sequences and Series": 5. Another comparison test

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Summary

• Let $N > 1$ be an integer, and consider a series $\sum_{n=1}^\infty a_n$. The series we get by removing the first $N-1$ terms, namely $$\sum_{n=N}^\infty a_n$$ is called a tail of the given series.
• Let $N > 1$ be an integer. The series $$\sum_{n=1}^\infty a_n$$ converges if and only if $$\sum_{n=N}^\infty a_n$$ converges.
  This could be shortened to "The series converges iff a tail of the series converges," or even just to the slogan that convergence depends on the tail.
• Limit Comparison Test
  Suppose $a_n \geq 0$ and $b_n \geq 0$. Then if $$\lim_{n \to \infty} \frac{a_n}{b_n} = L > 0,$$ the series $\sum_{n=1}^\infty a_n$ converges if and only if $\sum_{n=1}^\infty b_n$ converges.

Solution Manual of MOOCULUS-2 "Sequences and Series": 4. Alternating series

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Summary

• If $$\sum_{n=1}^\infty |a_n|$$ converges (i.e. absolutely convergent), then $$\sum_{n=1}^\infty a_n$$ converges (i.e. conditionally convergent).
• Suppose that $(a_n)$ is a decreasing sequence of positive numbers and $$\lim_{n\to\infty}a_n=0$$ Then the alternating series $$\sum_{n=1}^\infty (-1)^{n+1} a_n$$ converges.
• For an alternating series $$s_n=\sum_{n=1}^{\infty}(-1)^n\cdot a_n$$ the test steps:
  
  • If $$\lim_{n\to\infty}a_n\neq0$$ then it diverges;
  • If $$\lim_{n\to\infty}a_n=0$$ and $a_n$ converges, then it absolutely converges;
  • If $$\lim_{n\to\infty}a_n=0$$ and $a_n$ diverges, then it conditionally converges.

Solution Manual of MOOCULUS-2 "Sequences and Series": 3. Convergence tests

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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.

Solution Manual of MOOCULUS-2 "Sequences and Series": 2. Series

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Summary

• Suppose $(a_n)$ is a sequence with associated series $$\sum_{k=1}^\infty a_k$$ The sequence of partial sums associated to these objects is the sequence $$s_n = \sum_{k=1}^n a_k$$
• Consider the series $$\sum_{k=1}^\infty a_k$$ This series converges if the sequence of partial sums $$s_n = \sum_{k=1}^n a_k$$ converges. More precisely, if $$\lim_{n \to \infty} s_n = L$$ we then write $$\sum_{k=1}^\infty a_k = L$$ and say, "the series $\sum_{k=1}^\infty a_k$ converges to $L$."
  
  If the sequence of partial sums diverges, we say that the series diverges.
• A series of the form $$\sum_{k=0}^\infty a_0 \, r^k$$ is called a geometric series.
• Suppose $a_0 \neq 0$. Then for a real number $r$ such that $|r| < 1$, the geometric series $$\sum_{k=0}^\infty a_0\, r^k$$ converges to $\frac{a_0}{1-r}$.
  
  For a real number $r$ where $|r| \geq 1$, the aforementioned geometric series diverges.
• Consider the series $$\sum_{k=0}^\infty a_k$$ and suppose $c$ is a nonzero constant. Then $$\sum_{k=0}^\infty a_k$$ and $$\sum_{k=0}^\infty c\,a_k$$ share a common fate: either both series converge, or both series diverge.
  
  Moreover, when $$\sum_{k=0}^\infty a_k$$ converges, $$\sum_{k=0}^\infty c \, a_k = c \cdot \sum_{k=0}^\infty a_k$$
• Suppose $$\sum_{k=0}^\infty a_k$$ and $$\sum_{k=0}^\infty b_k$$ are convergent series. Then $$\sum_{k=0}^\infty (a_k+b_k)$$ is convergent, and
  $$\sum_{k=0}^\infty (a_k+b_k)=\left( \sum_{k=0}^\infty a_k\right)+\left(\sum_{k=0}^\infty b_k\right)$$
• If $$\sum_{k=0}^\infty a_k$$ converges then $$\lim_{n\to\infty}a_n=0$$
• Consider the series $$\sum_{k=0}^\infty a_k$$ If the limit $$\lim_{n\to\infty}a_n$$ does not exist or has a value other than zero, then the series diverges.
  
  We'll usually call this theorem the "$n^{\text{th}}$ term test."
• The series $$\sum_{n=1}^\infty {1\over n} = \frac{1}{1} +\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots$$ is called the harmonic series.
• Consider the series $$\sum_{k=0}^\infty a_k$$ Assume the terms $a_k$ are non-negative. If the sequence of partial sums $s_n = a_0 + \cdots + a_n$ is bounded, then the series converges.
• Comparison Test
  
  Suppose that $a_n$ and $b_n$ are non-negative for all $n$ and that, for some $N$, whenever $n \geq N$, we have $a_n \leq b_n$.
  
  If $$\sum_{n=0}^\infty b_n$$ converges, so does $$\sum_{n=0}^\infty a_n$$ If $$\sum_{n=0}^\infty a_n$$ diverges, so does $$\sum_{n=0}^\infty b_n$$
• Cauchy Condensation Test
  
  Suppose $(a_n)$ is a non-increasing sequence of positive numbers. The series $$\sum_{n=1}^\infty a_n$$ converges if and only if the series $$\sum_{n=0}^\infty \left( 2^n a_{2^n} \right)$$ converges.
• $p$-series Test
  $$\sum_{n=1}^{\infty} {1\over n^p}\ \begin{cases}\mbox{converges} & \mbox{when $p>1$}\\ \mbox{diverges} & \mbox{when $p\leq1$} \end{cases}$$

Solution Manual of MOOCULUS-2 "Sequences and Series": 1. Sequences

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Summary

• Suppose that $\left(a_n\right)$ is a sequence.
  To say that $\lim_{n\to \infty}a_n=L$ is to say that
  for every $\varepsilon>0$, there is an $N > 0$, so that whenever $n>N$, we have $|a_n-L| < \varepsilon$. If $\lim_{n\to\infty}a_n=L$ we say that the sequence converges.
  If there is no finite value $L$ so that $\lim_{n\to\infty}a_n = L$, then we say that the limit does not exist, or equivalently that the sequence diverges.
• Suppose $(a_n)$ is a sequence with initial index $N$, and suppose we have a sequence of integers $(n_i)$ so that $$N \leq n_1 < n_2 < n_3 < n_4 < n_5 < \cdots$$ Then the sequence $(b_i)$ given by $b_i = a_{n_i}$ is said to be a subsequence of the sequence $a_n$.
• If $(b_i)$ is a subsequence of the convergent sequence $(a_n)$, then $$\lim_{i \to \infty} b_i = \lim_{n \to \infty} a_n$$
• Suppose $(b_i)$ and $(c_i)$ are convergent subsequences of the sequence $(a_n)$, but $$\lim_{i \to \infty} b_i \neq \lim_{i \to \infty} c_i.$$
  Then the sequence $(a_n)$ does not converge.
• Squeeze Theorem:
  Suppose there is some $N$ so that for all $n > N$, it is the case that $a_n \le b_n \le c_n$. If $$\lim_{n\to\infty}a_n=\lim_{n\to\infty}c_n=L$$
  then $\lim_{n\to\infty}b_n=L$.
• $$\lim_{n\to\infty}|a_n|=0$$ if and only if $$\lim_{n\to\infty}a_n=0$$
• The sequence $a_n = r^n$ converges when
  $-1 < r \le 1$, and diverges otherwise. In symbols, $$\lim_{n\to\infty} r^n=\begin{cases}0& \mbox{if $-1 < r < 1$,} \\ 1& \mbox{if $r=1$, and} \\ \mbox{does not exist} & \mbox{if $r \leq -1$ or $r > 1$.}
  \end{cases}$$
• A sequence is called increasing (or sometimes strictly increasing) if $a_n < a_{n+1}$ for all $n$. It is called non-decreasing if $a_n\le a_{n+1}$ for all $n$.
  Similarly a sequence is decreasing (or, by some people, strictly decreasing) if $a_n > a_{n+1}$ for all $n$ and non-increasing if $a_n\ge a_{n+1}$ for all $n$.
• If a sequence is increasing, non-decreasing, decreasing, or non-increasing, it is said to be monotonic.
• A sequence $(a_n)$ is bounded above if there is some number $M$ so that for all $n$, we have $a_n\le M$.
  Likewise, a sequence $(a_n)$ is bounded below if there is some number $M$ so that for every $n$, we have $a_n\ge M$.
  If a sequence is both bounded above and bounded below, the sequence is said to be bounded.
• If the sequence $a_n$ is bounded and monotonic, then $\lim_{n \to \infty} a_n$ exists. In short, bounded monotonic sequences converge.