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

$$(a_n)$$ 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?

• Are you sure (c) is supposed to be $\sum \frac{a_n+1}{a_n}$ instead of $\sum \frac{a_{n+1}}{a_n}$? – Carl Schildkraut Feb 20 at 8:29
• 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. – Arthur Feb 20 at 8:30
• Yes fixed it – Abhay Feb 20 at 8:30
• counter example for d) $a_n = (2n)!$ – Abhay Feb 20 at 8:32
• Then assuming one has to be true, by elimination you've shown that (c) is the only possible one. – Keen-ameteur Feb 20 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.