1

When taking the limit of something, for example:

$$\lim\frac{1}{k+1}$$

as $k$ goes to infinity I was taught to multiply by $\frac{\frac{1}{k}}{\frac{1}{k}}$ to get

$$\lim\frac{\frac{1}{k}}{1+\frac{1}{k}}$$

where I was told to assume that $\frac{1}{k}$ goes to $0$ and I end up with

$$\frac{0}{1} = 0.$$

However, I'm confused as by the $p$-series test $\lim\frac{1}{k}$ is divergent as $p$ must be greater than $1$. But shouldn't $\frac{1}{k}$ be $0$ and therefore converge to $0$?

(All limits going to infinity)

Martin Sleziak
  • 53,687
NotSoTuringComplete
  • 41
  • careful: $\lim_{k\to\infty}\frac{1/k}{1+1/k}=\frac01=0$ but the expression $\lim_{k\to\infty} \frac01$ is not correct, you are taking limits twice. – Masacroso Dec 10 '17 at 03:32
  • Sorry, that $lim\frac{0}{1}$ shouldn't have had the $lim$ there. I'm confused as to why when I take $lim\frac{\frac{1}{k}}{1+\frac{1}{k}}$ that $\frac{1}{k}$ becomes $0$ but not when I take the limit of just $\frac{1}{k}$ – NotSoTuringComplete Dec 10 '17 at 03:36

2 Answers2

3

I think you may be confusing the distinction between the limit of the summand and the limit of the partial sums. I recently wrote up a big post on this for my sister here: Infinite amount of additions, finite sum?

Let $$s_n = \sum_{k = 1}^n \frac{1}{k}$$

$ \\ $ By definition,

$$\sum_{k = 1}^\infty \frac{1}{k} = \lim_{n \to \infty} s_n$$

Although $1/k \to 0$, $s_n \to \infty$

David Reed
  • 3,265
1

The $p$-series test applies to a series: an infinite sum. It tells you that $\sum_{k=1}^\infty \frac1k$ is divergent. It says nothing about the limit $\lim_{k \to \infty} \frac1k$.

Misha Lavrov
  • 142,276