Can an alternating series converge from below? 
Let $S_n=\sum_{k=0}^n (-1)^k a_k$ be convergent, with $a_k>0$ for all $k$. Can $S_n$ be always smaller than its limiting value?

I was thinking of something with non-monotone $a_k$, something like $1 - 1/2 + 1/3 - 1/6 + 1/5 - 1/15 + 1/10 \ldots$. But I am not sure how to continue this construction and whether it works.
 A: Yes. Let $a_{2k} = 1/2^k$ and $a_{2k+1} = 1/100^{k+1}$. Then $S_n$ in particular is absolutely convergent and the limiting value is $2 - 1/99$. One easily checks that $S_n$ is always smaller, intuitively because there is a big step of $1/2^k$ and the step of $-1/100^k$ does not pull $S_n$ down far enough. To see this, note that we have
$$S_{2k+1}-S_{2k-1} = 1/2^k - 1/100^{k+1} >0 \quad ∀ k$$
so $S_1,S_3,S_5,\dots$ is increasing and in particular $S_{2k+1} ≤ \lim_n S_n$.
For the even terms, since $1/2^{k+1} - 1/100^{k+2} ≥ 1/100^{k+1}$, $$S_{2k} = S_{2k+1} + 1/100^{k+1} ≤ S_{2k+1} + 1/2^{k+1}-1/100^{k+2} = S_{2k+3} ≤ \lim_n S_n$$
(note- the previous version of this answer was wrong but it has been fixed)
A: Choose any $x\in (1/2,1)$ and consider the series
$$2-1 +2x-x +2x^2-x^2+2x^3 -x^3 + \cdots.$$
Grouping in pairs shows this series converges to
$$ 1 + x+ x^2 + \cdots = \frac{1}{1-x}.$$
Looking at the partial sums, note that
$$S_{2n} = 1 + x + x^2 + \cdots + x^{n-1}<\frac{1}{1-x}$$
for all $n.$ For the odd partial sums we have
$$S_{2n+1} = 1+x + \cdots  + x^{n-1} + 2x^{n}.$$
Use the the formula for a finite geometric sum
and the condition $1/2 < x < 1$ to see $S_{2n+1}< \dfrac{1}{1-x}$ as well, and we're done.
