Convergence of Series in AP Calculus BC
This brief essay attempts to prove various theorems on convergence of series. The following theorems, better known as convergence tests, are presented in the AP Calculus BC curriculum.
We first review definitions related to convergence of series, then prove the Direct Comparison Test and the Limit Comparison Test; the two theorems serve as critical means of proving other ideas in this essay. Next, we prove the $n$-th Term Test and the Alternating Series Test, followed by the Ratio Test. Finally, we show results related to the Lagrange Error Bound of Taylor series.
Definitions
Definition 1. A sequence $a_k$ is a function with domain $\mathbb{N}$ and range $\mathbb{R}$.
Definition 2. A partial sum of a sequence $a_k$ is defined by yet another sequence
\[A_k := \sum^k_{n=0}a_n\]
Definition 3. A sequence $a_k$ converges to a limit $a$, i.e. $\lim_{k \to \infty} a_k=a$, iff
\[\forall \varepsilon \in \mathbb R ^+, \exists N \in \mathbb N, n \ge N \to |a-a_n|<\varepsilon\]If $a_k$ converges to a (finite) limit, it is called convergent.
Definition 4. Let $A_k$ be the partial sum of the sequence $a_n$. Then,
\[\sum ^\infty _{n=0}a_n := \lim_{k \to \infty} A_k\]
Definition 5. A sequence $a_k$ is nonincreasing if $\forall n, a_{n+1} \le a_n$ and nondecreasing if $\forall n, a_{n+1} \ge a_n$. If a sequence is either nondecreasing or nonincreasing, it is called monotone.
Theorem 1 (Monotone Sequence Theorem). A sequence $a_k \in \mathbb R$ is convergent iff it is bounded.
Moreover, $\lim a_k=\sup a_k$ if $a_k$ is nondecreasing, and $\lim a_k=\inf a_k$ if $a_k$ is nonincreasing.
The proof for the Monotone Sequence Theorem is omitted in this essay. Its proof is a direct consequence of the Completeness Axiom.
Note that we use $\lim a_k$ as shorthand for $\lim _{k \to \infty} a_k$, and $\sum a_k$ for $\sum^\infty _{k=n^*}a_k$.
Comparison Test Theorems
Theorem 2 (Direct Comparison Test). Given $0 \le a_n \le b_n$ for all $n \ge n^*$,
\[\sum^\infty _{n=n^*}b_n \text{ converges } \to \sum^\infty _{n=n^*}a_n \text{ converges }\]
Proof. Let $B_k, A_k$ be the partial sums of $b_n, a_n$, respectively (starting from $n=n^*$ to $n=k$).
Notice that $A_k$ and $B_k$ are nondecreasing (and thus monotone), as $A_{k+1}-A_k=a_k\ge 0$. Also, we are know that $A_k \le B_k$ because $b_n>a_n$ for all $n \ge n^*$.
It is given that $\lim B_k=B$. By the Monotone Sequence Theorem, $B=\sup B_k$, which is to say that for all $k$, $B_k<B$. Because $A_k<B_k<B$, $B$ is an upper bound of $A_k$. Again by the Monotone Sequence Theorem, $\lim A_k$ is convergent. Therefore $\sum^\infty _{n=n^*}a_n$ converges.
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Note the contrapositive of the above theorem:
\[\sum^\infty _{n=n^*}a_n \text{ diverges } \to \sum^\infty _{n=n^*}b_n \text{ diverges }\]Theorem 3 (Limit Comparison Test). If $\lim \frac{a_n}{b_n}=L>0$ where $a_n, b_n>0$, $a_n$ converges iff $b_n$ converges and $a_n$ diverges iff $b_n$ diverges.
Proof. Note the definition of convergence. We have that $\forall \varepsilon>0, \exists N \in \mathbb N,$
\[\left| \frac{a_n}{b_n} -L\right| <\varepsilon\] \[\iff L-\epsilon <a_n/b_n <L+\varepsilon\]Then, for sufficiently large $n$,
\[(L/2)b_n < a_n < (3L/2)b_n\]It is clear that $\sum b_n$ converges iff $\sum c\cdot b_n$ where $c$ is a constant unrelated to $n$.
The Direct Comparison Test concludes the proof, given the above relationship binding $a_n$ and $b_n$. Notice that this cannot be done if $\lim a_n/b_n=0$ or if the limit diverges.
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Other Theorems on Convergence
Theorem 4 ($n$th Term Test).
\[\lim a_k \ne 0 \to \sum a_k \text{ diverges}\]
Proof. Let $\lim a_k=L$. It is clear that for sufficiently large $n$, $a_n \ge L/2$. However, $\sum L/2$ is divergent. By the Direct Comparison Test, $\sum a_k$ must be divergent.
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Next, we consider the convergence of alternating series.
Definition 6. An alternating series is $\sum_k (-1)^ka_k$ where $\forall k, a_k>0$.
Claim 1. Let $A_k = \sum ^k _{n=0} (-1)^n a_n$ be the partial sum of an alternating series. We claim that if $a_n$ is positive and decreasing,
\[0<A_k<a_0\]
Proof. First consider $A_{2m}$. This sum can be written as
\[A_{2m}=a_0+\sum^m_{k=1}(-a_{2k-1}+a_{2k})\]where each $-a_{2k-1}+a_{2k}<0$ due to decreasingness. Therefore $A_{2m}<a_0$. Also,
\[A_{2m}=\sum_{k=0}^{m-1} (a_{2k}-a_{2k+1})+a_{2m}>0\]as $a_{2k}-a_{2k+1}>0$.
The proof for $A_{2m+1}$ is similar, and is left to the reader.
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Theorem 5 (Alternating Series Test). If $a_n$ is positive and decreasing with limit 0,
\[\sum^\infty _{n=0}(-1)^na_n \text{ converges}\]
Proof. Let $A_k := \sum ^k _{n=0} (-1)^n a_n$. We show that $\lim A_k$ is convergent.
Define $U_k=A_{2k}$. $U_k$ is decreasing (and thus monotone) because $U_{k+1}-U_k=A_{2k+2}-A_{2k}=a_{2k+2}-a_{2k+1}<0$. We also know that $\forall k, U_k>a_0$. Because $U_k$ is bounded from below and decreasing, it converges by the Monotone Sequence Theorem. It is trivial to show that if $U_k=A_{2k}$ converges, $A_k$ converges.
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Theorem 6 (Ratio Test).
\[\lim \left|\frac{a_{n+1}}{a_n}\right|=L<1 \to \sum |a_n| \text{ converges}\](i.e. that it converges absolutely) and
\[\lim \left|\frac{a_{n+1}}{a_n}\right|=L>1 \to \sum a_n \text{ diverges}\]
Proof. First, we show that $\lim |a_{n+1}/a_n|=L<1 \to \sum |a_k|$ is convergent. Let $b_k :=|a_k|$.
\[\lim b_{n+1}/b_n=L<1\] \[\iff \forall \varepsilon>0, \exists N \in \mathbb N, \ n \ge N \to L-\varepsilon<{b_{n+1}\over b_n}<L+\varepsilon\]We chose some $l \in (L,1)$ so that for all $n \ge N$, $b_{n+1}/b_n<l \iff b_{n+1}<lb_n$. It is easy to show, as $b_n$ is always nonnegative, that $b_n \le b_N\cdot l^{n-N}$. Because $0<L<l<1$, the infinite sum of the right-hand sequence is convergent (infinite geometric series). By the Direct Comparison Test, the $\sum b_n$ must be convergent as well. Therefore $\sum a_k$ is absolutely convergent, as $\sum |a_k|$ is convergent.
Second, we show that $\lim |a_{n+1}/a_n|=L>1$ implies that $\sum a_n$ is divergent. Choose $l \in (1,L)$ such that $\exists N \in \mathbb N, \forall \varepsilon>0, n \ge N \to |a_{n+1}/a_n-l|<\varepsilon$. Therefore $n \ge N$ then
\[|a_{n+1}|>l|a_n| \iff |a_n| > |a_N|\cdot l^{n-N}\]The sequence on the left-hand side is divergent. Therefore $\lim |a_n|$ and thus $\lim a_n$ is divergent. Thus, $\sum a_n$ is divergent.
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Langrange Error Bound for Taylor Series
Definition 7. The $n$-th taylor polynomial or taylor approximation of a function $f$ differentiable $n$ times is defined by
\[T_n(x)=\sum_{i=0}^n f^{(i)}(a)\frac{(x-a)^i}{i!}\]
Theorem 7 (Lagrange Error Bound). For a function $f$ differentiable $n+1$ times, $\exists z \in (a,x) \text{ or } (x,a)$ s.t.
\[|f(x)-T_n(x)|=|f^{(n+1)}(z)|\frac{(x-a)^{n+1}}{(n+1)!}\]
Proof. First assume $x>a$; the other case is similar. We use mathematical induction on $n$.
\[|f(x)-T_{n+1}(x)|=|f^{(n+2)}(u)|\frac{(x-a)^{n+2}}{(n+2)!}\] \[\textbf{LHS}=|f(x)-T_n(x)+T_n(x)-T_{n+1}(x)|\] \[=\left|\frac{|f^{(n+1)}(v)|}{(n+1)!}(x-a)^{n+1}+ \frac{|f^{(n+1)}(a)|}{(n+1)!}(x-a)^{n+1}\right|\]
- $n=0$: $T_0(x)=f(a)$. We state that $|f(x)-f(a)|=|f’(z)|(x-a)$ for some $z \in (a,x)$. This is true by the Mean Value Theorem.
- Assume that the statement for $n$ holds true. We consider $n+1$:
By the Mean Value Theorem,
\[\mathbf{RHS}=\frac{|f^{(n+1)}(x)-f^{(n+1)}(a)|}{x-a} \cdot (x-a)^{n+2}/(n+2)!\]for some $u$. This leaves the original equation equivalent to
\[||f^{(n+1)}(v)|-|f^{(n+1)}(a)||\] \[=\frac{1}{n+2}||f^{(n+1)}(x)|-|f^{(n+1)}(a)||\]for some $v$. To conclude the proof, we use Intermediate Value Theorem for $v$. The function $||f^{(n+1)}(v)|-|f^{(n+1)}(a)||$ $-\frac{1}{n+2}||f^{(n+1)}(x)|-|f^{(n+1)}(a)||$ is clearly positive for $v=x$, and negative for $v=a$. Therefore the function has a zero at some point $v \in (a,x)$.
Concluding the proof with the other case is left to the reader.
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