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$(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?

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  • $\begingroup$ 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 at 8:29
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    $\begingroup$ 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 at 8:30
  • $\begingroup$ Yes fixed it$$$$ $\endgroup$ – Abhay Feb 20 at 8:30
  • $\begingroup$ counter example for d) $a_n = (2n)!$ $\endgroup$ – Abhay Feb 20 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 at 8:33
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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.

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  • $\begingroup$ c) is true and all others are false. $\endgroup$ – Kavi Rama Murthy Feb 20 at 8:35

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