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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?

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All three terms are already meaningless for any fixed $n$. One can study the limit of a sequence if one really has one. – Pedro Tamaroff Sep 5 '13 at 3:25
@PeterTamaroff True, but does that necessarily imply they are meaningless as $n \rightarrow \infty$? Also, in light of your comment, I've edited the third limit so that it is meaningful for fixed $n$. – Colin T Bowers Sep 5 '13 at 3:29
I repeat: if you want to talk about $\lim\limits_{n\to\infty} s_n$, the $s_n$ have to be "something". In this case, at best, we can say $s_n=+\infty$. In the third case, the limit is $0$; since you're subtracting from $1$ the partial sums of $1+2^{-1}+\cdots$ that converge to it. – Pedro Tamaroff Sep 5 '13 at 3:31
@PeterTamaroff Thanks, I understand now :-) – Colin T Bowers Sep 5 '13 at 3:44
up vote 4 down vote accepted

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

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Thanks, that makes sense. – Colin T Bowers Sep 5 '13 at 3:40

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.

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Thanks, this is useful. I've given the answer tick to @Rebecca since you both provided the same answer but she posted first (by a small margin :-). Cheers. – Colin T Bowers Sep 5 '13 at 3:42

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