Sum from infinity to infinity

Question

How does one evaluate the following limits?

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

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

3. $\lim\limits_{n \to \infty} \sum\limits_{k=n}^\infty 2^{-k}$

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?

Answer 1

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

Answer 2

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.