Consider the convergent sequence 
with terms defined by
![\[ a_n = \frac{(-1)^{n+1}}{n}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-38a5d37ace1fbf352511c4a547737cfb_l3.png "Rendered by QuickLaTeX.com")
Let 
. Find the value of 
and values of 
such that 
for all 
for each of the following values of 
:
-
,
-
,
-
,
-
,
-
.
First, we know
![\[ \lim_{n \to \infty} \frac{(-1)^{n+1}}{n} \leq \lim_{n \to \infty} \left| \frac{(-1)^{n+1}}{n} \right| = \lim_{n \to \infty} \frac{1}{n} = 0 \quad \implies \quad L= 0. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-1989370ebd27d9ddf83e84fba6e59d67_l3.png "Rendered by QuickLaTeX.com")
So then we have,
Thus, if 
then for every we have 
. We compute for the given values of as follows:
- implies
.
- implies
.
- implies
.
- implies
.
- implies
.
I find myself nitpicking here: When you passed to the absolute value you must have assumed the limit L is zero, which is obvious in this case. But beyond knowing the limit in advance Apostol didn’t give any tools for actually determining convergence. What little I know of analysis it seems something like all convergent subsequences sharing the same limit would work, or perhaps cauchy criterion? It just seems that Apostols exposition of sequences was less then comforting; but maybe it will be covered later.
Greetings!
There are of course alternative to solve this exercise. You can take
and show that for 
we indeed have 
for 
arbitrary. Thus proving 
as 
using only the definition of convergence. From 
you can also create the required table of 
. (I use 
instead of 
).
Yeah but it should be
?