Criteria for convergent sequence (Baby Rudin Theorem 3.22)

Question

Theorem 3.22 of Rudin's Principles of Mathematical Analysis says:

The series $\sum a_{n}$ of (real or) complex numbers converges iff for every $\varepsilon > 0$, there is an integer $N$ such that $$ \left| \sum_{k=n}^m a_{k} \right| \leq \varepsilon $$ if $m\geq n \geq N.$

My doubt is: The condition is that the sequence of partial sums {$s_{n}$} is Cauchy, i.e for every $\varepsilon>0$ there is an integer $N$ such that $$ \left|s_{m}-s_{n} \right|<\varepsilon $$ if $m \geq n \geq N.$

But $\left|s_{m}-s_{n} \right|$ would be $\left|\sum_{k=n+1}^m a_{k}\right|$, isn't it?

How come the summation starts from $n$ in the theorem? How are the two summations equivalent?

Answer 1

It's irrelevant. Replacing $N$ with $N+1$ will not change how the statement works. The point is that for each $\varepsilon>0$ there exists some $N$ such that from then on all tails are less than $\varepsilon$.

Answer 2

Here's the detailed to arrive there. As you said, if we regard sequence $\{s_n\}$ with $s_n = \sum_{k=1}^n a_k$, then the Cauchy criterion for sequence (Theorem 3.11) will translate to :

$\sum a_n$ converges (i.e $s_n$ converges to $s$ ) $\iff$ $\{s_n\}$ Cauchy, that is for any $\epsilon >0$, there are integer $N$ such that for $m,n \geq N$ $$ |s_m-s_n|=\left|\sum_{k=n+1}^m a_k \right| < \epsilon. $$

Now we have to make a way such that the expression from theorem 3.22 holds from this. That is

$\sum a_n$ converges $\iff$ for any $\epsilon >0$, there are integer $N$ such that for $m,n \geq N$ , $\left|\sum_{k=n+1}^m a_k \right| < \epsilon$

$\iff$

for any $\epsilon >0$, there are integer $N$ such that for $m,n \geq N$, $\left|\sum_{k=n}^m a_k \right| \leq \epsilon$.

To show the $\implies$ part, just note that
$$\sum_{k=n}^m a_k = a_n +\sum_{k=n+1}^m a_k, \quad \text{and} \quad a_n= s_n-s_{n-1}. $$ So we have inequality \begin{align} \left|\sum_{k=n}^m a_k \right| &= \left| a_n +\sum_{k=n+1}^m a_k \right| \leq |a_n| + \left|\sum_{k=n+1}^m a_k \right| = |s_n-s_{n-1}| + \left|\sum_{k=n+1}^m a_k \right| \\ &= |s_n - s + s-s_{n-1}| + \left|\sum_{k=n+1}^m a_k \right| \\ &\leq |s_n - s| + |s-s_{n-1}| + \left|\sum_{k=n+1}^m a_k \right| \end{align} Because $\{s_n\}$ converge, then given $\epsilon/3 >0$, we have integer $N_1$ such that for any $n-1 \geq N_1$ we have $|s_{n-1}-s|<\epsilon/3$. This also implies $|s_n-s|<\epsilon/3$ because $n>n-1\geq N_1$. We know also that we have integer $N_2$ such that for any $m,n \geq N_2$ implies $\left|\sum_{k=n+1}^m a_k \right| \leq \epsilon/3$. Set $N = \text{max} \{N_1,N_2\}$, then by above inequality for all $m,n \geq N$ implies \begin{align} \left|\sum_{k=n}^m a_k \right| &\leq |s_n - s| + |s-s_{n-1}| + \left|\sum_{k=n+1}^m a_k \right| < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon . \end{align} We can show the $\impliedby$ part by the same way.

Answer 3

I think this question has answers that are either to complicated or too vague.

Here is my take on it, which expands on the idea from the accepted answer, but shows in detail how indexes can be manipulated here:

Let's proove:

$$ \forall \epsilon > 0 \quad \exists n_0 \in \mathbb{N} \quad \forall m \geq n \geq n_0 \quad | \sum_{k=n}^m a_k | < \epsilon $$

iff

$$ \forall \epsilon > 0 \quad \exists n_1 \in \mathbb{N} \quad \forall m \geq n \geq n_1 \quad | \sum_{k=n+1}^m a_k | < \epsilon $$

Note how I marked two numbers $ n_0, n_1 $ to avoid any confusion.

Proof "down":

Take any $ \epsilon > 0 $ By assumption we get $ n_0 $. Take $ n_1 = n_0 $. Take any $ m \geq n \geq n_1 $. Then:

$$ m \geq n_1 = n_0 $$ $$ n + 1 > n \geq n_1 = n_0 $$

If $ m = n $ we have two show that $ | \sum_{k=m + 1}^m a_k | = 0 < \epsilon $, which is true.

If $ m > n $ then $ m \geq n + 1 $, so we have $ m \geq n + 1 \geq n_0 $, which we now can plug into our assumption to get:

$$ |\sum_{k=n+1}^m a_k | < \epsilon $$

which is exactly what we wanted.

Proof "up":

Take any $ \epsilon > 0 $. By assumption we get $n_1 $. Take $ n_0 = n_1 + 1 $. Take any $ m \geq n \geq n_0 $.

Then:

$$ m \geq n_0 = n_1 + 1 > n_1 $$

$$ n \geq n_0 = n_1 + 1 $$

and thus:

$$ n - 1 \geq n_1 $$

Obviously $ m \geq n - 1 \geq n_1 $, so we can plug these numbers into bottom assumption and get:

$$ | \sum_{k=n-1+1}^m a_k | < \epsilon $$

which is what we wanted.

Answer 4

By Cauchy criterion for sequence (Theorem 3.11), $s_n$ converges if and only if
$$\forall \varepsilon > 0 \quad\exists N_0\in \mathbb{N}\quad \forall m>n-1\geq N_0, \quad| s_m-s_{n-1} | \leq \varepsilon$$ That is,

$$\forall \varepsilon > 0 \quad\exists N_0\in \mathbb{N}\quad \forall m\geq n\geq N_0+1, \quad| \sum_{k=n}^m a_k | \leq \varepsilon$$ Let $N_0+1=N$, we have $$\forall \varepsilon > 0 \quad\exists N\in \mathbb{N}\quad \forall m\geq n\geq N, \quad| \sum_{k=n}^m a_k | \leq \varepsilon$$