Convergence of 1/2^(1+infinitesimal) If $a$ is an infinitesimal or a non-standard number or a number strictly larger then 0 and smaller then all positive real numbers. What is 
$\sum_{n=1}^{\infty} \frac{1}{2^{1+a}} $
$\lim_{x\to \infty} (1+a)^x$
$\sum_{n=1}^{\infty} a $
 A: By the transfer principle, any theorem of standard analysis is a theorem of non-standard analysis.
For example, $\lim_{x \to +\infty} b^x = +\infty$ when $b > 1$ is a theorem of standard analysis. Thus, its transfer is a theorem of non-standard analysis. Since $1+a > 1$, we have $\lim_{x \to +\infty} (1+a)^x = +\infty$.
Of course, if $H$ is a positive transfinite number, then $\lim_{x \to H} (1+a)^x$ is not $+\infty$, it is $(1+a)^H$. This value may or may not be transfinite, depending on whether $H$ is large enough (This is closely related to indeterminate forms), But it is definitely smaller than $+\infty$.
The same principle applies to your other examples.
EDIT: with the update (and since Kermit seems to want more comment), we know that $\sum_{n=1}^{+\infty} 1/n^{1+a} = \zeta(1+a)$ and $\sum_{n=1}^{+\infty} a = +\infty$ for all $a > 0$. These statements remain true in the non-standard model.
A: Until kermit says what he means, we can only speculate.  
Suppose $F$ is an ordered field, strictly larger than $\mathbb R$.  Then there is an element $a \in F$ such that: $a>0$ but $a<r$ for all real $r>0$.  Equivalently: $a>0$ but $a<1/n\;$ for all $n \in \mathbb N$.  (Of course $\mathbb R$ contains a subset we naturally call $\mathbb N$.)  I will consider a question in the same vein as kermit's questions, but--I think--more elementary:  
$$\lim_{x\to\infty} xa$$
Since things like this are not conventionally defined in $F$, but only in $\mathbb R$, we will have to say what it means.  There is more than one possibility.  Here is one:  
(1)  Let $b \in F$.  We say $\lim_{x\to \infty} xa = b$ iff for every positive $\epsilon \in F$, there exists $N \in \mathbb N$ such that for all $n \in \mathbb N$, if $n \ge N$, then $|na-b|<\epsilon$.  If there is such an element $b$, then we say $\lim_{x\to\infty} xa$ exists.  
(1')  We say $\lim_{x\to \infty} xa = \infty$ iff for every $k \in F$, 
there exists $N \in \mathbb N$ such that for all $n \in \mathbb N$, if $n \ge N$, then $na > k$.  
Here is another:  
(2)  Let $b \in F$.  We say $\lim_{x\to \infty} xa = b$ iff for every positive $\epsilon \in F$, there exists $X \in F$ such that for all $x \in F$, if $x \ge X$, then $|xa-b|<\epsilon$.  If there is such an element $b$, then we say $\lim_{x\to\infty} xa$ exists.  
(2')  We say $\lim_{x\to \infty} xa = \infty$ iff for every $k \in F$, 
there exists $X \in F$ such that for all $x \in F$, if $x \ge X$, then $xa > k$.  
I numbered them (1) and (2) for reference, but there are still more possibilities...  
