# Let $p_n$ be the sequence defined by $p_n=\sum_{k=1}^n\frac{1}{k}$. Show that $p_n$ diverges even though $\lim_{n\to\infty}(p_n-p_{n-1})=0$ [duplicate]

I have tried this as :

$$p_n=\sum_{k=1}^n\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n-1}+\frac{1}{n}$$

$$p_{n-1}=\sum_{k=1}^{n-1}\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n-1}$$

$$(p_n-p_{n-1})=\frac{1}{n}$$

$$\lim_{n\to\infty}(p_n-p_{n-1})=\lim_{n\to\infty}\frac{1}{n}=0$$

But dont know how to show $p_n$ diverges?

## marked as duplicate by Thomas Andrews, David Mitra, Martin, Amzoti, Stefan HansenJun 19 '13 at 16:35

• You mean $\sum_{k=1}^n \frac{1}{k}$. Because $\sum_{k=1}^n \frac 1 n=1$. – Thomas Andrews Jun 19 '13 at 15:52
• Anyway, just use the Cauchy condensation test. – Hagen von Eitzen Jun 19 '13 at 15:58
• @ThomasAndrews Thank you for your source. – time Jun 19 '13 at 16:02

Using some idea from the comments, with $\,a_n:=\frac1n\;$:

Proof 1: Cauchy's Condensation test:

$$2^na_{2^n}=\frac{2^n}{2^n}=1\;,\;\text{and since }\;\;\sum_{n=1}^\infty 1\;\;\text{diverges so does our series}$$

Proof 2: Integral test:

$$\int\limits_1^\infty\frac{dx}x=\left.\lim_{b\to\infty}\log x\right|_1^b=\lim_{b\to\infty}\log b=\infty\implies \;\text{ also our series diverges}$$

Proof 3: Subsequence of sequence of partial sums:

$$a_{2^n}=1+\frac12+\ldots+\frac1{2^n}=$$

$$=\left(1+\frac12\right)+\left(\frac13+\frac14\right)+\left(\frac15+\ldots+\frac18\right)+\ldots+\left(\frac1{2^{n-1}+1}+\frac1{2^{n-1}+2}+\ldots+\frac1{2^n}\right)\ge$$

$$\ge\underbrace{ \frac12+\frac12+\ldots+\frac12}_{n\text{ times}}=\frac n2\xrightarrow[n\to\infty]{}\infty$$

and thus our series diverges.

• Thank you very much. thank you, all. – time Jun 20 '13 at 3:11

$$\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$$

$$\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}$$

Generalising, $$\frac{1}{2^n-2^{n-1}+1}+\dots +\frac{1}{2^n}>\frac{1}{2^n}+\dots +\frac{1}{2^n}\text{ (2^{n-1} terms)}$$

Hence, we can add an infinite number of $\frac{1}{2}$. The sum thus diverges.

This is a standard exercise. How many terms does it take to get a sum greater than 1/2. How many more to get another 1/2. Keep doing this and you will see why the series diverges.

The above hint will lead you to the classical proof by Oresme and Cauchy. There are other proofs by Mengoli, Jakob Bernoulli, Johann Bernoulli and Euler, as well as several modern ones.

Here is my favorite variant of the problem: One day a calculus student started off on a magic road that was sixteen meters long, traveling at a constant rate of one meter per second. At the end of each second the road instantaneously and uniformly grew by sixteen meters. Does the student ever get to the end of the calculus road? If so, when?

hmm, seems that you just need to prove that the series $\displaystyle\sum_{k=1}^{\infty}\frac 1 k$ diverges (that will be enough when $n\to\infty$). I'll give the explanation that my lecturer gave when I took calculus I:

Let's develop the harmonic series a little bit: $$\sum\limits_{k=1}^{\infty}\frac 1 k=1+\frac 1 2 +\frac 1 3+\frac 1 4 +\frac 1 5+\dots.$$ Let's devide the sum into "blocks" in size of $2^i(i\in\mathbb N\cup 0)$ in the following way: $$\sum\limits\frac 1 k=(1)+(\frac 1 2+ \frac 1 3)+(\frac 1 4+\frac 1 5+\frac 1 6+\frac 1 7)+(\frac 1 8+\frac 1 9+\frac 1 {10}+\frac 1 {11}+\frac 1 {12}+\frac 1 {13}+\frac 1 {14}+\frac 1 {15})$$Notice that each of the blocks is $\ge \frac 1 2$ so we got: $$\sum \frac 1 k\geq \frac 1 2+\frac 1 2+\frac 1 2+\frac 1 2+\frac 1 2+\frac 1 2+\frac 1 2+\frac 1 2+\frac 1 2+\dots$$ or in other words the sum is bigger than infinite blocks in size of $\frac 1 2$ so we can conlude: $$\sum\limits_{k=1}^{\infty}\frac 1 k= \infty$$ $\Rightarrow$converges. I hope I didn't have any mistakes and that it solves the issue.

Recall the inequality $e^x>1+x \forall x>0$

Now Consider $$e^{p_n}=e^{\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)}$$ $$=e\cdot e^{\frac{1}{2}}\cdot e^{\frac{1}{3}}\cdots e^{\frac{1}{n}}$$ $$>\left(1+1\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdots\left(1+\frac{1}{n}\right)$$ $$=(2)\left(\frac{3}{2}\right)\left(\frac{4}{3}\right)\cdots\left(\frac{n+1}{n}\right)$$ $$=n+1$$

Thus, taking $\ln$ on both sides (inequality maintains as $\ln$ is an increasing function), $$p_n>\ln(n+1)$$ and $\ln(n+1)$ is unbounded as $n\uparrow$ and hence $p_n$ diverges.