$(a_n) $ is a sequence of positive real numbers. The series $\sum a_n$ will converge if

(a) $\sum a_n^2$ converges.

(b)$\sum \frac{a_n}{2^n}$ converges

(c)$\sum \frac{a_{n+1}}{a_n}$ coverges

(d)$\sum \frac{a_n}{a_{n+1}}$ converges

a) can't be true, counter example : $\sum\frac{1}{n^2}$ converges but not $\sum \frac1n$

b) can't be true, counter example : $\frac{n}{2^n}$ converges but not $\sum n$

I can't decide between c and d. I think c might be true by taking $a_n = \frac{1}{(2n)!}$

also I think taking $a_n = (2n)!$ will disprove d also. So is c the correct option?

  • $\begingroup$ Are you sure (c) is supposed to be $\sum \frac{a_n+1}{a_n}$ instead of $\sum \frac{a_{n+1}}{a_n}$? $\endgroup$ – Carl Schildkraut Feb 20 '19 at 8:29
  • 2
    $\begingroup$ You can't prove any of these to be true by giving an example. You can prove they're not true by giving a counterexample, but that's it. $\endgroup$ – Arthur Feb 20 '19 at 8:30
  • $\begingroup$ Yes fixed it$$$$ $\endgroup$ – Abhay Feb 20 '19 at 8:30
  • $\begingroup$ counter example for d) $a_n = (2n)!$ $\endgroup$ – Abhay Feb 20 '19 at 8:32
  • $\begingroup$ Then assuming one has to be true, by elimination you've shown that (c) is the only possible one. $\endgroup$ – Keen-ameteur Feb 20 '19 at 8:33

If $\sum \frac {a_{n+1}} {a_n}$ converges then $\frac {a_{n+1}} {a_n} \to 0$ so $\sum a_n$ converges by ratio test. If $\sum \frac {a_n} {a_{n+1}}$ converges then $\frac {a_n} {a_{n+1}} \to 0$ and $\frac {a_{n+1}} {a_n} \to \infty $, so ratio test tells you that $\sum a_n$ diverges.

| cite | improve this answer | |
  • $\begingroup$ c) is true and all others are false. $\endgroup$ – Kavi Rama Murthy Feb 20 '19 at 8:35

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.