Can one construct a convergent series out of a non-decreasing Cauchy sequence in $\mathbb{R}$ If we have a non-decreasing Cauchy sequence of non-negative real numbers $\{x_n\}$, will the sequence of partial sums $\{S_n\}_{n=1}^{\infty}$ where $S_n= x_1 + x_2 +..+x_n$ converges?
 A: Suppose $S_n$ converges, it implies $x_n\to 0$
Since we have $x_n\ge 0$ then the sequence will be globally decreasing (e.g $\forall n\ \exists n_0>n\ \mid \forall i> n_0,\,x_i\le x_{n_0}$)
But the sequence can be non-decreasing while being globally decreasing. It suffice for instance to interleave two decreasing sequences.
$x_{2n} = \dfrac{1}{n^\alpha}$ and $x_{2n+1}=\dfrac{2}{n^\alpha}$ for instance.
We have $(x_{2n})_n\searrow$ and $(x_{2n+1})_n\searrow$ but $x_{2n+1}>x_{2n}$ so  $(x_n)_n$ is non-decreasing.

Depending on the value of $\alpha$ the partial sum $S_n$ will converge or diverge.
So the answer for the TITLE is YES, you can construct such series, take $\alpha=2$ for instance.
And the answer for the wording below is NOT NECESSARILY, take $\alpha=1$ for instance.
A: No. All that being a Cauchy sequence tells you is that the terms eventually get very close to each other. However, the could do this in the vicinity of $5$, for example $a_n = 5 + \dfrac 1n$. Then the sum is $\ge 5+5+5 + \cdots$, which diverges.
A: No. All convergent sequences are Cauchy, but  that doesn't mean that a series of positive terms which tend to $0$ (hence the sequence of these terms is Cauchy) is itself Cauchy, as this would imply that a series with positive terms converges if and only if its general tends to $0$, which false in $\mathbf R$ (but is true in the field of $p$-adic numbers).
