Lecture - Analysis MT22, X

Let $\sum _ {k\ge 1} a _ k$ be a series and $(s _ n)$ be the series of $n$-th partial sums. What does it mean for the series to converge?

$(s _ n)$ converges.

Let $\sum _ {k\ge 1} a _ k$ be a series and $(s _ n)$ be the series of $n$-th partial sums. How can you recover $a _ n$ from $s _ n$?

\[a_n = s_n - s_{n-1}\]

Given that $a _ n = s _ n - s _ {n-1}$ (where $s _ n$ is the $n$-th partial sum), what is the necessary and sufficient condition for $a _ n \ge 0$?

$s _ n$ is monotonic increasing.

What theorem/statement relates the convergence of $\sum _ {k \ge 1} a _ k$ and $a _ k \to 0$?

If $\sum _ {k\ge 1} a _ k$ converges, then $a _ k \to 0$ as $k \to \infty$.

It is true that if $\sum _ {k\ge 1} a _ k$ converges, then $a _ k \to 0$ as $k \to \infty$. What’s an example of the converse not being true?

\[a_k = \frac{1}{k}\]

How can you prove the theorem that if $\sum _ {k\ge 1} a _ k$ converges, then $a _ k \to 0$ as $k \to \infty$?

Consider rewriting $a _ n$ as $s _ n - s _ {n-1}$.

To prove that $\sum _ {k\ge 1} \frac{1}{k}$ diverges, why is it enough to show $s _ n = \sum^n _ {k \ge 1} \frac{1}{k}$ is not Cauchy?

Not being cauchy implies divergence.

 What $n$ and $m$ do you take in Cauchy’s convergence criterion for proving the divergence of the harmonic series?

\[n = 2^{p+1}\text{, }m = 2^p\]

What’s the point of taking $n = 2^{p+1}\text{, }m = 2^p$ in Cauchy criterion for $s _ n = \sum^n _ {k \ge 1} \frac{1}{k}$?

You can show that $ \vert s _ n - s _ m \vert $ will always be greater than $\frac{1}{2}$.

If you have real series $\sum _ {k\ge 1} a _ k$ and $\sum _ {k\ge 1} b _ k$, what is the condition and statement for the simple comparison test?

Condition: $$

\exists c > 0 \text{ s.t } 0 \le a_k \le Cb_k \text{ for } k \ge 1

\[Statement:\]

\sum_{k\ge 1} b_k \text{ convergent} \implies \sum_{k \ge 1}a_k \text{ convergent}

$$

What does it mean for a series $\sum _ {k \ge 1}a _ k$ to be absolutely convergent?

\[\sum_{k \ge 1} |a_k| \text{ converges}\]

What theorem relates absolute convergence and normal convergence for a real or complex series $\sum _ {k \ge 1} a _ k$?

\[\sum_{k \ge 1} |a_k| \text{ converges} \implies \sum_{k \ge 1} a_k \text{ converges}\]

What does Cauchy’s criterion simplify into when considering the convergence of a series $\sum _ {k \ge 1} a _ k$?

\[\forall \epsilon > 0 \text{ } \exists N \text{ s.t. } n > m > N \implies |\sum^n_{k = m+1} a_k| < \epsilon\]