3
$\begingroup$

We are able to show, using the Cauchy criterion (using sum from $n$ to $2n$) that $nx_n \to 0$ Explicitly this is $0<nx_{2n}<\displaystyle \sum_{i=n}^{2n}x_n$ and the result follows from squeeze rule. My question is can we do better?

Intuitively, it feels like we should since $\frac{n\log n}{n^k} \to 0$ for $k>1$. Unfortunately, I am struggling to prove this, Any hints would be much appreciated.

One idea I had was to say that if $\displaystyle \sum_{i=N}^{n} x_i > \sum_{i=N}^{n} \frac{1}{n^k}$ for $k>1$ and for all sufficiently large $n$ then we could somehow compare it to the harmonic series. Could we not say the tail of the series could be made as large as we want this way and so divergence would be inevitable?

$\endgroup$
  • 1
    $\begingroup$ @Winther "decreasing" $\endgroup$ – Gabriel Romon May 20 at 20:17
  • $\begingroup$ How about $x_n = \frac{1}{n \log n}$? $\endgroup$ – sudeep5221 May 20 at 20:20
  • $\begingroup$ @sudeep5221 $\sum \frac{1}{n\log n}$ doesn't converge. $\endgroup$ – Thomas Andrews May 20 at 20:21
  • $\begingroup$ Doesn't converge, use Cauchy Condensation test @sudeep5221 $\endgroup$ – user3184807 May 20 at 20:21
  • $\begingroup$ Yeah my bad I forgot that condition. $\endgroup$ – sudeep5221 May 20 at 20:21
7
$\begingroup$

Let $N_k = 2^{k^2}$ and define $(x_n)$ by $x_1 = 1$ and $x_n = \frac{1}{N_k \log N_k}$ if $N_{k-1} < n \leq N_k$ and $k \geq 1$. Then $(x_n)$ is positive and (weakly) decreasing. Also,

$$ \sum_{n=1}^{\infty} x_n = 1 + \sum_{k=1}^{\infty} \frac{N_k - N_{k-1}}{N_k \log N_k} \leq 1 + \sum_{k=1}^{\infty} \frac{1}{k^2\log 2} < \infty. $$

On the other hand, $(n \log n) x_n = 1$ for infinitely many $n$, and so, it does not converge to $0$.

$\endgroup$
  • 4
    $\begingroup$ Note: the sequence thus defined is only non-increading, not decreasing. However, "perturbing" it by a ridiculously negligible amount (e.g., $y_i = x_i + 1/2^i$) will fix that, leaving the argument intact. $\endgroup$ – Clement C. May 20 at 20:30
  • $\begingroup$ Wow! That's quite impressive, thanks! $\endgroup$ – user3184807 May 20 at 20:38
3
$\begingroup$

This holds when additionally $nx_n$ decreases to $0$.

Indeed let us state the following lemma: if $(b_n)$ is a sequence that decreases to $0$, $\sum_n \frac{b_n}{n} <\infty \implies b_n=o\left( \frac 1{\log n}\right)$.

To prove the lemma, note that by condensation, $\sum_n b_{2^n}$ converges, hence $b_{2^n}=o\left(\frac 1n\right)$ and $b_n=o\left( \frac 1{\log n}\right)$.

Applying the lemma with $b_n = nx_n$ yields the claim.

$\endgroup$
-2
$\begingroup$

Hint: show that the series $$ \sum_{n=2}^\infty \frac{1}{n (\ln n)(\ln \ln n)}$$ is divergent. From this it follows that $x_n$ must decrease faster, if $\sum_n x_n$ is to be convergent.

$\endgroup$
  • 2
    $\begingroup$ That... is not a proof, nor an argument that actually works. $\endgroup$ – Clement C. May 20 at 20:31
  • $\begingroup$ I said it's a hint. $\endgroup$ – Adam Latosiński May 20 at 20:32
  • 2
    $\begingroup$ Then it's a hint that doesn't work. $\endgroup$ – Clement C. May 20 at 20:32
  • $\begingroup$ Considering sequences of that form cannot be made to work I dont think $\endgroup$ – user3184807 May 20 at 20:33
  • $\begingroup$ Except the statement isn't true, so the hint doesn't do any good. $\endgroup$ – Thomas Andrews May 20 at 20:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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