Proofs - Analysis I MT22
Basic properties
Sequences
Series
Prove the limit form of the comparison test:
Let $(a _ k)$ and $(b _ k)$ be real sequences of positive terms, and assume that there is $L \in (0, \infty)$ such that $\frac{a _ k}{b _ k} \to L$ as $k \to \infty$. Then $\sum a _ k$ converges if and only if $\sum b _ k$ converges.
Let $(a _ k)$ and $(b _ k)$ be real sequences of positive terms, and assume that there is $L \in (0, \infty)$ such that $\frac{a _ k}{b _ k} \to L$ as $k \to \infty$. Then $\sum a _ k$ converges if and only if $\sum b _ k$ converges.
- First prove the normal form of the comparison test, which follows from $a _ k$ being a monotone increasing sequence
- Then consider what $\frac{a _ k}{b _ k} \to L$ tells you by taking $\varepsilon = L/2$ and use the regular comparison test for both directions.
(analysis i, page 90)
Prove the alternating series test:
Let $(u _ k)$ be a real sequence, and consider the series $\sum^\infty _ {k=1} (-1)^{k-1} u _ k$. If $u _ k \ge 0$ for $k \ge 1$, $(u _ k)$ is decreasing, and $u _ k \to 0$ as $k \to \infty$, then the series converges.
Let $(u _ k)$ be a real sequence, and consider the series $\sum^\infty _ {k=1} (-1)^{k-1} u _ k$. If $u _ k \ge 0$ for $k \ge 1$, $(u _ k)$ is decreasing, and $u _ k \to 0$ as $k \to \infty$, then the series converges.
- Group terms together in different ways in order to see that both the $s _ {2n}$ and $s _ {2n+1}$ series of partial sums are convergent by the monotone sequence theorem.
(analysis i, page 88)
Prove the ratio test:
Let $(a _ k)$ be a real sequence of positive terms. Assume that $\frac{a _ {k+1}\,}{a _ k}$ converges as $k \to \infty$, say to a limit $L$. Then if $0 \le L < 1$, then $\sum a_k$ converges and if $L > 1$, then $\sum a _ k$ diverges.
Let $(a _ k)$ be a real sequence of positive terms. Assume that $\frac{a _ {k+1}\,}{a _ k}$ converges as $k \to \infty$, say to a limit $L$. Then if $0 \le L < 1$, then $\sum a_k$ converges and if $L > 1$, then $\sum a _ k$ diverges.
Need to handle two cases seperately, but in both we compare with a geometric series. In the case where $0 \le L < 1$, this is $\alpha = \frac{1+L}{2}$ with $\varepsilon = \alpha - L$, in the other case it’s $\alpha = 2$.
(analysis i, page 92)
Prove the integral test:
Let $f : [1, \infty) \to \mathbb R$ be a function. Assume that
- $f$ is nonnegative
- $f$ is decreasing
- $\int^{k+1} _ k f(x) \text d x$ exists for each $k \ge 1$
Then let $s _ n = \sum^n _ {k=1} f(k)$ and $I _ n = \int^n _ 1 f(x) \text d x$. Then
- $\sigma _ n = s _ n - I _ n$ converges and $0 \le \lim _ {n \to \infty} \sigma _ n \le f(1)$
- $s _ n$ converges iff $I _ n$ converegs.
Let $f : [1, \infty) \to \mathbb R$ be a function. Assume that
- $f$ is nonnegative
- $f$ is decreasing
- $\int^{k+1} _ k f(x) \text d x$ exists for each $k \ge 1$ Then let $s _ n = \sum^n _ {k=1} f(k)$ and $I _ n = \int^n _ 1 f(x) \text d x$. Then
- $\sigma _ n = s _ n - I _ n$ converges and $0 \le \lim _ {n \to \infty} \sigma _ n \le f(1)$
- $s _ n$ converges iff $I _ n$ converegs.
- Show that $\sigma _ n$ is bounded below and decreasing, by adding up all the inequalities like $f(n) \le \int^n _ {n-1} f(x) \text d x \le f(n-1)$ to get $0 \le \sigma _ n \le f(1)$ and then mess with $\sigma _ {n+1} - \sigma _ n$ to show it’s negative.
- Then the other result follows from the AOL.
(analysis i, page 96).
Radius of convergence
Prove that power series converge within their radius of convergence $R$:
Let $\sum c _ k z^k$ be a power series with radius of convergence $R$. Then if $R > 0$ and $ \vert z \vert < R$ then $\sum c_k z^k$ converges absolutely and hence converges. If $ \vert z \vert > R$ then $\sum c _ k z^k$ diverges.
Let $\sum c _ k z^k$ be a power series with radius of convergence $R$. Then if $R > 0$ and $ \vert z \vert < R$ then $\sum c_k z^k$ converges absolutely and hence converges. If $ \vert z \vert > R$ then $\sum c _ k z^k$ diverges.
- Two cases parts of the proof, absolute convergence within radius of convergence and then divergence outside it
- Absolute convergence within radius of convergence: There is some $S$ less than the radius of convergence, and let $\varepsilon = R - S$ to find a $\rho$ less than or equal to $R$ where the series converges absolutely. Then use the comparison test.
- Divergence outside radius of convergence: Use contradiction. Can bound the terms being summed by $M$, and then can find some $\rho$ where $R < \rho < \vert z \vert $ where $0 \le \vert c _ k \rho^k \le \vert c _ k z^k \vert \left \vert \frac \rho z \right \vert ^k \le M \left \vert \frac \rho z\right \vert ^k$ which converges as it’s a geometric series with common ratio less than $1$.