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

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

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

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

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?
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 $$\lim_{n \rightarrow \infty} \sum_{k=n}^\infty \frac{1}{2^k}=0.$$
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?