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Example

We show $\sum\ \frac{ 1 }{ n } = +\infty$.

Proof [expand]

$\sum\ \frac{ 1 }{ n } \geq \int_{ 1 }^{ n+1 } \frac{ 1 }{ x } \d x = \log_e(n+1)$

Since $\limu{ n }{ \infty } \log_e(n+1) = + \infty , \sum\ \frac{ 1 }{ n } = + \infty$.

Note

We use integral test in this case. Notice how we choose the lower limit. It’s pretty arbitrary as long as it is larger than $1$ and meet our need. Think about it, why can we do this.

Example

We show $\sum \frac{ 1 }{ n^2 }$ converges.

Proof [expand]

$\sum_{ k=1 }^{ n } \frac{ 1 }{ k^2 } \leq 1 + \int_{ 1 }^{ n } \frac{ 1 }{ x^2 } \d x = 2 - \frac{ 1 }{ n } < 2$.

$\therefore \sum \frac{ 1 }{ n^2 }$ converges.

Definition: Alternating Series

Let $a_n >0 $, $\sum_{ n = 1 }^{ \infty } (-1)^{n+1} a_n$ is alternating.

any convergent series has $a_n \to 0 $.

Suppose $a_n > 0 $ and $a_n \to 0 $

suppose $a_n > a_{n+1} \forall n$ i.e. decreasing.

Theorem

$\sum \frac{ 1 }{ n^p } $ converges if and only if $p > 1 $.

Definition: Alternating Series

$\sum (-1)^{n+1} a_n $ is called an alternating series.

Theorem : Alternating Series Theorem

If $a_1 \geq a_2 \geq … \geq a_n \geq … \geq 0 $ and $\lim a_n = 0 $, then the alternating series $\sum\ (-1)^{n +1} a_n $ converges. Moreover, the partial sums $s_n = \sum_{ k=1 }^{ n } (-1)^{k+1} a_k $ satisfy $\abs{ s - s_n } \leq a_n $for all $n $.

Proof [expand]

We need to show that the sequence $(s_n)$ converges. Note that the subsequence $(s_{2n}) $ is increasing because $s_{2n+2} - s_{2n} = + a_{2n+1} - a_{ 2n + 2 } \geq 0 $. Similarly, the subsequence $(s_{2n-1}) $ is decreasing since $s_{2n+1} - s_{2n-1} = a_{2n+1} - a_{2n} \leq 0 $. We claim

$$s_{2m} \leq s_{2n+1} \forall m, n \in \N \tag{1}$$

First note that $s_{2n} \leq s_{2n+1} $ for all $n $, because $s_{2n+1} -s_{2n} = a_{2n+1} \geq 0$. If $m \leq n $, then $(1)$ holds because $s_{2n+1} \geq s_{2m+1} \geq s_{2m} $. Thanks to $(1) $, we see that $(s_{2n}) $ is an increasing subsequence of $(s_n) $ bounded above by each odd partial sum, and $(s_{2n+1}) $ is a decreasing subsequence of $(s_n) $ bounded below by each even partial sum. By theorem 10.2, these subsequences converge, say to $s $ and $t $. Now

$$t - s = \limu{ n }\infty{ s_{2n+1}} - \limu{ n} \infty { s_{2n}} = \limu{ n} \infty {(s_{2n+1} - s_{2n})} = \limu{ n} \infty { a_{2n+1}} = 0 $$

so $s = t$. It follows that $\lim_n s_n = s$.