Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question already has an answer here:

I don't know if my proof is correct. Let be

\begin{eqnarray} H_n &=& 1 + \dfrac{1}{2} + \dfrac{1}{3} + \dotsb + \dfrac{1}{n} \\ &+& \\ H_n &=& \dfrac{1}{n} + \dfrac{1}{n-1} + \dfrac{1}{n-2} + \dotsb + 1\\ &\parallel& \\ 2H_n &=& \dfrac{n+1}{n}+ \dfrac{n+1}{2(n-1)} + \dfrac{n+1}{3(n-2)} + \dotsb + \dfrac{n+1}{k(n-k+1)} + \dotsb + \dfrac{n+1}{n} \\ &\parallel& \\ 2H_n &=& (n+1) \sum_{k=1}^n \dfrac{1}{k(n-k+1)} \\ &\parallel& \\ H_n &=& \dfrac{(n+1)}{2} \sum_{k=1}^n \dfrac{1}{k(n-k+1)}\\ \end{eqnarray}

Let's say $b_n = \sum_{k=1}^n \dfrac{1}{k(n-k+1)}$. The sequence $b_n$ is strictly increasing, so only two things are going to happen (this part I'm not sure) :

If $b_n$ is convergent then $\lim\limits_{n \rightarrow \infty} b_n >0$ and $\lim\limits_{n \rightarrow \infty} H_n = \lim\limits_{n \rightarrow + \infty} \frac{(n+1)}{2} . b_n = + \infty$

If $b_n$ diverges then $\lim\limits_{n \rightarrow \infty} b_n = + \infty$ and $\lim\limits_{n \rightarrow \infty} H_n = \lim\limits_{n \rightarrow + \infty} \frac{(n+1)}{2} . b_n = + \infty$

Therefore, harmonic series is divergent.

share|improve this question

marked as duplicate by Nate Eldredge, Bruno Joyal, T. Bongers, Stefan4024, Sujaan Kunalan Nov 22 '13 at 3:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
That is correct. My fault –  jon jones Jun 27 '12 at 21:37
    
Since this is homework, I think you have to show more precisely why $b_{n}$ is strictly increasing. Observe that it depends on $n$ both in the index of the summation and in the fraction. –  madprob Jun 27 '12 at 21:50
    
You say that the sequence $(b_n)$ is strictly increasing. I don't see that, is false. –  André Nicolas Jun 27 '12 at 21:51
    
@AndréNicolas And I'm thinking it goes to $0$. It can't be all nice and rainbows to prove the harmonic series diverges! –  Pedro Tamaroff Jun 27 '12 at 21:58

2 Answers 2

Given that André showed your proof is not correct, I have an alternate idea to give you.

We know by the comparison test, that if the series $$\sum_{n=1}^\infty a_n$$ diverges and $$\lim\limits_{n \to \infty} \dfrac{b_n}{a_n}=\ell >0 $$ then

$$\sum_{n=1}^\infty b_n$$

diverges too.

But from elementary calculus, we know that

$$\mathop {\lim }\limits_{x \to 0^+} \frac{{\log \left( {1 + x} \right)}}{x} = 1$$

This means that letting $x=1/n$

$$\mathop {\lim }\limits_{n \to +\infty } \frac{{\log \left( {1 + \frac{1}{n}} \right)}}{{\frac{1}{n}}} = 1$$

Does $${\sum\limits_{n = 1}^\infty {\log \left( {1 + \frac{1}{n}} \right)} }$$ converge or diverge?


Propted by Hennings comment, I add the following:

Cauchy's Condensation Test

If $a_n$ is a decreasing sequence of positive terms, then $$\sum_{n>0} a_n$$ converges if and only if $$\sum_{n \geq 0} 2^n a_{2^n}$$ does, and

$$\sum_{n>0} a_n \leq \sum_{n \geq 0} 2^n a_{2^n} \leq 2 \sum_{n >0} a_n $$

(This can be proven rather easily)

share|improve this answer
    
An even more elementary idea is $$\sum_{n=2^k+1}^{2^{k+1}} \frac 1n \ge \sum_{n=2^k+1}^{2^{k+1}} \frac1{2^{k+1}} = \frac 12$$ for all $k$, hence $\sum_{n=1}^{2^k} \frac 1n \ge \frac k2$. –  Henning Makholm Jun 27 '12 at 23:15
    
@HenningMakholm That is called Cauchy's Condensation test. In particular, it works off perfectly in this case. $$\sum a_{n}$$ converges iff $a_n$ is a decreasing positive sequence, and $$\sum 2^n a_{2^n}$$ does. In this case condensation gives $$\sum_{n>0} 1$$ (Very cute outcome!) –  Pedro Tamaroff Jun 27 '12 at 23:18

Unfortunately, the proof breaks down, since the sequence $(b_n)$ cannot be strictly increasing. In fact, $(b_n)$ has limit $0$.

For if $(b_n)$ were increasing, then by your formula the sum of the first $n$ terms of the harmonic series would grow at least like a constant times $n+1$. But it only grows logarithmically. So for large $n$, $b_n$ is approximately $a_n=\frac{2\log n}{n+1}$.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.