1
$\begingroup$

I will now show that $\sum \frac{1}{2k}$ and $\frac{1}{2k+1}$ both diverge.

$\exists \ \epsilon > 0 \ \forall N \in \mathbb{N}$ so that for $m>n>N$:

$0\le |\sum_{k=n}^{m} \frac{1}{2k+1} | \le |\sum_{k=n}^{m} \frac{1}{2k}| < |\sum_{k=n}^{m} \frac{1}{k}| > \epsilon $ for some $n>N$

Tell me if this is formally correct. Please.

  • 2
    Notice that your chain of inequalities does not actually relate the series of interest to $\epsilon$ in any way.2011-10-28
  • 2
    It will be nicer if you try to explain in words instead of a chain of symbols. How exactly are you planning to obtain the conclusion?2011-10-28
  • 0
    For all n>N there is a epsilon so that the harmonic series diverges. And since the harmonic series diverges and the other two are always smaller than it , by the comparison test they also diverge.2011-10-28
  • 1
    You have that backwards. If the series were always larger than the harmonic, then they would also diverge. Very much not vice versa.2011-10-28
  • 0
    I don't see where I wrote that the series of interest are larger than the harmonic series. It is easy to see that they are always smaller since $|\frac{1}{2k+1}| < |\frac{1}{2k}| < |\frac{1}{k}| \ \forall k\in \mathbb{N}$. How to formally write that , by the comparison test and with the cauchy criterion it follows that because $\frac{1}{k}$ diverges, so do the other two??2011-10-28
  • 0
    I said IF they were larger, you could use the comparison test. Since they are not, you can't.2011-10-28
  • 0
    Is this correct: Assume the series and the harmonic series do converge, then from $0\le \sum | \frac{1}{2k+1} | \le \sum | \frac{1}{2k} \le \sum |\frac{1}{k} | < \epsilon $ it would follow that all three are convergent, but since $\sum |\frac{1}{k}|$ does not converge, neither do the other two .2011-10-28
  • 1
    @VVV No, that is not correct. For instance, consider $0<\sum \frac 1 {k^2}<\sum \frac 1 k$, where $\sum \frac 1 {k^2}$ converges but $\sum \frac 1 k$ does not. Take a look at AMPerrine's method again.2011-10-28

3 Answers 3