Zondrina said:
Wait a second... a thought occurred to me a moment ago. I'm not really sure it has any merit (since pas got off), but I will post it and see what anyone who views this may think.
Suppose without loss of generality for the duration of this question that n > m.
Let the sequences of partial sums be defined as : ##s_n = \sum_{k=1}^{n} a_k## and ##s_m = \sum_{k=1}^{m} a_k##
We know that ##\sum a_n## converges which implies ##a_n → 0## as ##n → ∞##.
This is true.
the sequences of partial sums also converge.
This is the definition of convergence of an infinite series.
So by the C.C, we want to show ##\forall ε>0, \exists N \space | \space n > m > N \Rightarrow |s_n - s_m| < ε##
No. We already know this: we are given that \sum a_n converges. Thus by definition the sequence of partial sums converges, and the Cauchy criterion then says that for all \epsilon > 0, there exists N \in \mathbb{N} such that if n > m \geq N then |s_n - s_m| < \frac12\epsilon.
Since m > n, we know that ##s_n## contains all the terms of ##s_m## plus extra left over term(s). To find these other terms we consider the following :
Let k = n - m so that k + m = n, then we have :
##|s_{m+k} - s_m| ≤ |s_k| = |a_{m+1} + ... + a_{n}| ≤ a_{m+1} + ... + a_{n}##
The first relation is true (at least under the assumption that the a_n are non-negative and non-increasing) but unhelpful. The next relation is false: s_k = a_1 + \cdots + a_k. The final relation is false unless all the terms are non-negative.
It ought to be obvious that if m < n then
<br />
\sum_{k=1}^n a_k = \sum_{k=1}^m a_k + \sum_{k=m+1}^n a_k<br />
so that
|s_n - s_m| = \left|\sum_{k=1}^n a_k - \sum_{k=1}^m a_k\right| = \left|\sum_{k=m+1}^n a_k\right| = |a_{m+1} + \cdots + a_{n}| = a_{m+1} + \cdots + a_{n}
with the last equality following because the terms are non-negative.
Now from the hint, we know that ##a_{m+1} + ... + a_{n} ≥ (n-m)a_n##.
So, putting the hint together with the above, we have that for all \epsilon > 0 and n > m sufficiently large,
<br />
(n - m)a_n \leq a_{m+1} + \dots + a_n = |s_n - s_m| < \textstyle \frac12 \epsilon<br />
from which it follows that
<br />
na_n < \textstyle\frac12 \epsilon + ma_n<br />
which was what the first part of the question asked you to prove.
Also, for showing ##na_n → 0##, fix m=0 as suggested.
You can't fix m = 0. The assumption we're operating under is that n > m \geq N where N is given by the Cauchy criterion as above.
Since ##\sum a_n## converges, once again as stated before, ##a_n → 0## as ##n → ∞##. So multiplying ##a_n## by an integer will not affect the result. Thus ##na_n → 0## as ##n → ∞##.
I'm not sure what you're trying to say here. If you're saying that for fixed m, ma_n\to 0 since a_n \to 0 then you are right, but that only gets you
<br />
0 \leq \lim_{n \to \infty} na_n \leq \textstyle\frac12 \epsilon
and you need to explain why that requires that na_n \to 0. If you're saying that na_n \to 0 if a_n \to 0 (which if you're assuming m = 0 might be the case) then that is simply false.
As for the example take : ##a_n = \frac{(-1)^n}{n}##.
This doesn't satisfy the condition that na_n \to 0.
Finding examples is almost impossible if you don't know where to look, so I'll just suggest that you look at (n\ln n)^{-1} for n \geq 2 (it really doesn't matter what you define a_1 to be as long as a_1 \geq (2\ln 2)^{-1} so the sequence is non-increasing).
The point of the question is that you've proven that if \sum a_n converges and the terms are non-negative and non-increasing, then na_n \to 0. However the last part of the question shows that the converse does not hold: the fact that na_n \to 0 does not imply that \sum a_n converges.
