Reading: Read sections 13, 14, 15.
Exercises:
Show for bounded sequences and . Hint: First show
Proof. Let and consider . As , . Similarly, . So,
so that is an upper bound for the set . In particular, this means that
Taking limits as of both sides preserves the inequality, so that
and applying the definition of yields,
as claimed.
This question is worth double points. Let be any sequence of non-negative real numbers. Let be defined by,
Show
Hint: For the last inequality, show first that implies
Show that if exists, then converges and
Give an example of a sequence so that converges but does not.
Proof. We have trivially that , so we need only prove the first and last inequalities. First we will prove the inequality:
Consider and any . Consider any . We will bound from above:
where we have used that (1) as all are nonnegative, (2) , and (3) that for all we have . So,
As . This upper bound holds for all , so we may conclude that
As this relationship holds for all , taking limits of either side as (we may do this as increasing preserves the property that ) we obtain,
so that .
Taking limits then as we get that
as desired.
We now show the first inequality, which uses similar techniques but is slightly different. We take , any , and consider arbitrary . Then we examine a lower bound for ;
where we have used that (1) as all are nonnegative, and (2) that for all we have . Notice now that , so that
This lower bound holds for all so we have proved that
Similar to before, taking limits as yields that
and taking limits as yields finally that
as desired.
Proof. If exists then and so by (a) .
Here is one example:
Let . Then diverges, but what of ? For even we can see that and odd we have that . This means that and and we conclude that .
Suppose \(\sum a_n = A\) and \(\sum b_n = B\) where \(A, B \in \mathbb R\).
Prove that \(\sum(a_n+b_n) = A+B\).
Prove that \(\sum ka_n = kA\) for any \(k \in \mathbb R\).
Is it the case that \(\sum a_nb_n = AB\)? Explain your reasoning.
Let \(\alpha_N = \sum_{n=1}^{N} a_n\) and \(\beta_N = \sum_{n=1}^{N} b_n\). Then
\[ \lim_{N\to\infty} \sum_{n=1}^N {a_n + b_n} = \lim_{N\to\infty}{\alpha_N + \beta_N} = \lim_{N\to\infty}{\alpha_N} + \lim_{N\to\infty}{\beta_N} = A+B \]
Let \(\alpha_N = \sum_{n=1}^{N} a_n\). Then
\[ \lim_{N\to\infty} \sum_{n=1}^N {k a_n} = \lim_{N\to\infty}{k \alpha_N} = k\lim_{N\to\infty}{\alpha_N} = kA \]
No. Consider \((a_n) = (1, 3, 0, \cdots)\) and \((b_n) = (1, 2, 0, \cdots)\). Then \(A = 4\) and \(B = 3\) but \(\sum a_nb_n = 1+6 = 7 \ne 12\).
For each of the following, determine whether the given series converges and justify your answer.
Converges by root test; \(\lim \vert a_n \vert^{1/n} = 1/2 < 1\)
Converges by ratio test; \(\lim \left\vert \frac{a_{n+1}}{a_n} \right\vert = \lim \left(\frac{n}{n+1}\right)^n = 1/e < 1 \)
Converges by comparison to \(\sum 1/3^n\), which is a convergent geometric series.