Sum from infinity to infinity How does one evaluate the following limits?

*

*$\lim\limits_{n \to \infty} \sum\limits_{k=n}^\infty (1)$


*$\lim\limits_{n \to \infty} \sum\limits_{k=n}^\infty k^{-1}$


*$\lim\limits_{n \to \infty} \sum\limits_{k=n}^\infty 2^{-k}$
Do all three limits evaluate to $0$? If so, why? Perhaps only the third limit evaluates to $0$, while the first two are undefined. Again, why? @PeterTamaroff pointed out that the first two limits are undefined for fixed $n$, but does that necessarily imply they are undefined in the limit?
 A: Consider the sequence $$\left(\sum_{k=n}^\infty 1\right)_{n \geq 1}.$$ This sequence is $$(\infty,\infty,\infty,\ldots).$$  Hence, if we wish to define a limit of this sequence, it should be $$\lim_{n \rightarrow \infty} \sum_{k=n}^\infty 1=\lim_{n \rightarrow \infty} \infty=\infty.$$
The second example is resolved in the same way (since the series is also divergent).
In the third case, the sequence is $$\left(\sum_{k=n}^\infty \frac{1}{2^k}\right)_{n \geq 1}.$$  Since $$\sum_{k \geq 1} \frac{1}{2^k}=1,$$ the sequence is $$(1,\tfrac{1}{2},\tfrac{1}{4},\ldots)$$ so $$\lim_{n \rightarrow \infty} \sum_{k=n}^\infty \frac{1}{2^k}=0.$$
A: As limits are usually defined, you can only take $\lim_{n \to \infty}$ of an actual sequence of real numbers, so the first two limits are in fact meaningless, while the third evaluates to $0$ since $\frac{1}{2^{n-1}} \to 0$.
However, it is possible to discuss convergence in the extended real line $\mathbb{R} \cup \{-\infty,\infty\} = [-\infty, \infty]$.  In this framework, observe that the sequences of partial sums for the first two sums both approach $\infty$, so both sums are equal to $\infty$ for any fixed $n$.  Then the first and second limits are the limit of the sequence $\infty, \infty, \infty, \dots$, which converges to $\infty$.  The third limit is still $0$ as it was before.
A: Although your second sum

2) $\lim_{n \rightarrow \infty} \sum_{k=n}^\infty k^{-1}$

converges to 0, it is intresting that 
$\lim_{n \rightarrow \infty} \sum_{k=n}^{2n} k^{-1} = \ln 2$
looking for other cases that base index starts from some infinity?
