2
$\begingroup$

Let $(a_n)$ be an unbounded strictly increasing sequence of positive real numbers and let $x_k=\frac{a_{k+1}-a_{k}}{a_{k+1}}$. Then I want to find the correct option (s).

  1. For all $n\geq m, \sum\limits^{n}_{k=m}x_k>1-\frac{m}{n}$
  2. For all $n\geq m, \sum\limits^{n}_{k=m}x_k>\frac{1}{2}$
  3. $\sum\limits^{\infty}_{k=1}x_k$ converges to a finite limit
  4. $\sum\limits^{\infty}_{k=1}x_k$ diverges to $\infty$

My observation says that option 2 is false and so is 3. But I couldn't conclude anything for 1 and 4. Please help!

$\endgroup$
1
$\begingroup$

Note first that perhaps you have $a_k>0$. If $a_{k+1}=0$, you have a problem.

A) For 1), you can take $a_k=\log(ka)$, $a\geq 2$, and say that your inequality (for fixed $n,m$) cannot hold for large $a$.

B) For 4), if first $x_k$ do not $\to 0$, then your series $x_k$ do not converge. Suppose now $x_k\to 0$. Show that the (positive) series $y_k=-\log(1-x_k)$ is divergent, and use $-\log(1-x)\sim x$ if $x\to 0$ to finish.

$\endgroup$
  • $\begingroup$ Yes, I am sorry $a_k>0$ is given. $\endgroup$ – Anupam Dec 31 '14 at 17:09

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