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True or false: if $a_n$ is any decreasing sequence of positive real numbers and $b_n$ is any sequence of real numbers converges to 0, then $a_n\over b_n$ diverges.

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if $a_n=e^{-n} $ and $b_n=\frac {1}{n^2} $ then

$(a_n) $ is a decreasing sequence of positive numbers and $\lim_{+\infty}b_n=0$

but

$$\frac {a_n}{b_n}=n^2e^{-n}\to 0.$$

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Hint: Take $a_n=b_n=\frac{1}{n}$

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  • $\begingroup$ ya, that's what I did, but I wasn't sure if it was correct. A sequence of decreasing positive numbers should converge to zero. thanks $\endgroup$ – Jasmine Jul 24 '17 at 0:56
  • $\begingroup$ @Jasmine "A sequence of decreasing positive numbers should converge to zero"... that statement is not at all true. $c_n=1+\frac{1}{n}$ is a decreasing sequence of positive real numbers which converges to $1$ instead. It just so happens that the example tattwamasi gives converges to zero in this specific case. $\endgroup$ – JMoravitz Jul 24 '17 at 0:58
  • $\begingroup$ @Jasmine: A sequence of decreasing positive numbers need not converge to zero. $c_n=1+\frac 1n$ is decreasing and positive but converges to $1$. This example meets your requirements on $a_n,b_n$ and it is constant so it converges. $\endgroup$ – Ross Millikan Jul 24 '17 at 0:59
  • $\begingroup$ thanx for the correction $\endgroup$ – Jasmine Jul 24 '17 at 1:24
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Hint: Take $a_n = b_n$ for all $n$

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