What is $\lim_{n \to \infty} \sum_{k=n}^\infty 1/k$? Random question but I was just wondering this and can't seem to find an answer anywhere. I want to say that it is infinity but I'm not quite sure.
Edit: Let me actually put it another way, since this is the question I'm really wondering: is it true that $\inf_{n \in \mathbb{N}} \sum_{k=n}^\infty 1/k = \infty$ ?
 A: The expression you are considering doesn't make sense since the harmonic series diverges.
For finite sum we have that
$$\sum_{k=n}^N \frac1k =\sum_{k=1}^{N} \frac1k-\sum_{k=1}^{n-1} \frac1k=H_N-H_{n-1}\sim\log N-\log (n-1)$$
A: You are asking for 
$$\lim\left\{\sum_{k=1}^\infty 1/k, \sum_{k=2}^\infty 1/k, \sum_{k=3}^\infty 1/k, \ldots\right\}$$
which would correspond to the meaningless:
$$\lim\left\{\infty, \infty, \infty, \ldots\right\}$$
So there is no real answer to the question. Although you could modify some standard definitions for limits and call this $\infty$.
A: Let me put it this way. Assume there is a natural number $n$ such that $\sum_{k=n}^\infty\frac{1}{k}$ is finite. Then there are a finite number of natural numbers less than $n$, and since adding on a finite number of terms to a convergent sequence will not affect convergence, we have that $\sum_{k=1}^{n-1}\frac{1}{k}+\sum_{k=n}^\infty\frac{1}{k}$ converges. But we see that this is a contradiction, as this is equal to $\sum_{k=1}^\infty\frac{1}{k}$, which obviously diverges. Hence, for all $n\in\mathbb{N}$, $\sum_{k=n}^\infty\frac{1}{k}$ diverges. What does this tell you about the following set?
$$S=\left\{n\in\mathbb{N}:\sum_{k=n}^\infty\frac{1}{k}<\infty\right\}$$
