Counterexamples related to a convergent positive series












0












$begingroup$



Let ${a_n}$ be a sequence such that $a_n > 0$ for all $n ≥ 1$ and $sum_1^infty$$a_n$
converges.



Give
counterexamples to the following claims where $b_n$ =
$a_{n+1}$/$a_n$



(a) $a_n ≤ 1$ for all $n ≥ 1$.



(b) The sequence ${a_n}$ is non-increasing.



(c) $lim_{nto infty}b_n$ exists.



(d) If $lim_{nto infty}b_n$ exists, then $lim_{nto infty}b_n < 1$.



(e) The sequence ${b_n}$ is bounded.



(f) If $limsup_{nto infty}b_n$ exists, then $limsup_{nto infty}b_nleq 1$.




My attempt: $2/n^3$ works for (a) and (d). I would really appreciate help for the other counter-examples.










share|cite|improve this question











$endgroup$












  • $begingroup$
    I came up with an answer for part (b) as well, so I'm now just looking for help on (c), (e), and (f). My proposed solution isn't a counterexample for these cases, I believe.
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 6:38










  • $begingroup$
    You should add these to your post.
    $endgroup$
    – xbh
    Dec 11 '18 at 6:44










  • $begingroup$
    $a_n := frac{1}{2^n}$ satisfies $lim_n b_n = frac{1}{2}$, so (c),(e),(f) are satisfied
    $endgroup$
    – mathworker21
    Dec 11 '18 at 6:47










  • $begingroup$
    @mathworker21 no, we have to come up with counter-examples for these statements.
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 6:50










  • $begingroup$
    @childishsadbino my bad. idk, just have $a_n$ alternate between $frac{1}{2^n}$ and $frac{1}{n^2}$, or something
    $endgroup$
    – mathworker21
    Dec 11 '18 at 6:51


















0












$begingroup$



Let ${a_n}$ be a sequence such that $a_n > 0$ for all $n ≥ 1$ and $sum_1^infty$$a_n$
converges.



Give
counterexamples to the following claims where $b_n$ =
$a_{n+1}$/$a_n$



(a) $a_n ≤ 1$ for all $n ≥ 1$.



(b) The sequence ${a_n}$ is non-increasing.



(c) $lim_{nto infty}b_n$ exists.



(d) If $lim_{nto infty}b_n$ exists, then $lim_{nto infty}b_n < 1$.



(e) The sequence ${b_n}$ is bounded.



(f) If $limsup_{nto infty}b_n$ exists, then $limsup_{nto infty}b_nleq 1$.




My attempt: $2/n^3$ works for (a) and (d). I would really appreciate help for the other counter-examples.










share|cite|improve this question











$endgroup$












  • $begingroup$
    I came up with an answer for part (b) as well, so I'm now just looking for help on (c), (e), and (f). My proposed solution isn't a counterexample for these cases, I believe.
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 6:38










  • $begingroup$
    You should add these to your post.
    $endgroup$
    – xbh
    Dec 11 '18 at 6:44










  • $begingroup$
    $a_n := frac{1}{2^n}$ satisfies $lim_n b_n = frac{1}{2}$, so (c),(e),(f) are satisfied
    $endgroup$
    – mathworker21
    Dec 11 '18 at 6:47










  • $begingroup$
    @mathworker21 no, we have to come up with counter-examples for these statements.
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 6:50










  • $begingroup$
    @childishsadbino my bad. idk, just have $a_n$ alternate between $frac{1}{2^n}$ and $frac{1}{n^2}$, or something
    $endgroup$
    – mathworker21
    Dec 11 '18 at 6:51
















0












0








0





$begingroup$



Let ${a_n}$ be a sequence such that $a_n > 0$ for all $n ≥ 1$ and $sum_1^infty$$a_n$
converges.



Give
counterexamples to the following claims where $b_n$ =
$a_{n+1}$/$a_n$



(a) $a_n ≤ 1$ for all $n ≥ 1$.



(b) The sequence ${a_n}$ is non-increasing.



(c) $lim_{nto infty}b_n$ exists.



(d) If $lim_{nto infty}b_n$ exists, then $lim_{nto infty}b_n < 1$.



(e) The sequence ${b_n}$ is bounded.



(f) If $limsup_{nto infty}b_n$ exists, then $limsup_{nto infty}b_nleq 1$.




My attempt: $2/n^3$ works for (a) and (d). I would really appreciate help for the other counter-examples.










share|cite|improve this question











$endgroup$





Let ${a_n}$ be a sequence such that $a_n > 0$ for all $n ≥ 1$ and $sum_1^infty$$a_n$
converges.



Give
counterexamples to the following claims where $b_n$ =
$a_{n+1}$/$a_n$



(a) $a_n ≤ 1$ for all $n ≥ 1$.



(b) The sequence ${a_n}$ is non-increasing.



(c) $lim_{nto infty}b_n$ exists.



(d) If $lim_{nto infty}b_n$ exists, then $lim_{nto infty}b_n < 1$.



(e) The sequence ${b_n}$ is bounded.



(f) If $limsup_{nto infty}b_n$ exists, then $limsup_{nto infty}b_nleq 1$.




My attempt: $2/n^3$ works for (a) and (d). I would really appreciate help for the other counter-examples.







sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 7:03









Robert Z

101k1070143




101k1070143










asked Dec 11 '18 at 6:25









childishsadbinochildishsadbino

1148




1148












  • $begingroup$
    I came up with an answer for part (b) as well, so I'm now just looking for help on (c), (e), and (f). My proposed solution isn't a counterexample for these cases, I believe.
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 6:38










  • $begingroup$
    You should add these to your post.
    $endgroup$
    – xbh
    Dec 11 '18 at 6:44










  • $begingroup$
    $a_n := frac{1}{2^n}$ satisfies $lim_n b_n = frac{1}{2}$, so (c),(e),(f) are satisfied
    $endgroup$
    – mathworker21
    Dec 11 '18 at 6:47










  • $begingroup$
    @mathworker21 no, we have to come up with counter-examples for these statements.
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 6:50










  • $begingroup$
    @childishsadbino my bad. idk, just have $a_n$ alternate between $frac{1}{2^n}$ and $frac{1}{n^2}$, or something
    $endgroup$
    – mathworker21
    Dec 11 '18 at 6:51




















  • $begingroup$
    I came up with an answer for part (b) as well, so I'm now just looking for help on (c), (e), and (f). My proposed solution isn't a counterexample for these cases, I believe.
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 6:38










  • $begingroup$
    You should add these to your post.
    $endgroup$
    – xbh
    Dec 11 '18 at 6:44










  • $begingroup$
    $a_n := frac{1}{2^n}$ satisfies $lim_n b_n = frac{1}{2}$, so (c),(e),(f) are satisfied
    $endgroup$
    – mathworker21
    Dec 11 '18 at 6:47










  • $begingroup$
    @mathworker21 no, we have to come up with counter-examples for these statements.
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 6:50










  • $begingroup$
    @childishsadbino my bad. idk, just have $a_n$ alternate between $frac{1}{2^n}$ and $frac{1}{n^2}$, or something
    $endgroup$
    – mathworker21
    Dec 11 '18 at 6:51


















$begingroup$
I came up with an answer for part (b) as well, so I'm now just looking for help on (c), (e), and (f). My proposed solution isn't a counterexample for these cases, I believe.
$endgroup$
– childishsadbino
Dec 11 '18 at 6:38




$begingroup$
I came up with an answer for part (b) as well, so I'm now just looking for help on (c), (e), and (f). My proposed solution isn't a counterexample for these cases, I believe.
$endgroup$
– childishsadbino
Dec 11 '18 at 6:38












$begingroup$
You should add these to your post.
$endgroup$
– xbh
Dec 11 '18 at 6:44




$begingroup$
You should add these to your post.
$endgroup$
– xbh
Dec 11 '18 at 6:44












$begingroup$
$a_n := frac{1}{2^n}$ satisfies $lim_n b_n = frac{1}{2}$, so (c),(e),(f) are satisfied
$endgroup$
– mathworker21
Dec 11 '18 at 6:47




$begingroup$
$a_n := frac{1}{2^n}$ satisfies $lim_n b_n = frac{1}{2}$, so (c),(e),(f) are satisfied
$endgroup$
– mathworker21
Dec 11 '18 at 6:47












$begingroup$
@mathworker21 no, we have to come up with counter-examples for these statements.
$endgroup$
– childishsadbino
Dec 11 '18 at 6:50




$begingroup$
@mathworker21 no, we have to come up with counter-examples for these statements.
$endgroup$
– childishsadbino
Dec 11 '18 at 6:50












$begingroup$
@childishsadbino my bad. idk, just have $a_n$ alternate between $frac{1}{2^n}$ and $frac{1}{n^2}$, or something
$endgroup$
– mathworker21
Dec 11 '18 at 6:51






$begingroup$
@childishsadbino my bad. idk, just have $a_n$ alternate between $frac{1}{2^n}$ and $frac{1}{n^2}$, or something
$endgroup$
– mathworker21
Dec 11 '18 at 6:51












2 Answers
2






active

oldest

votes


















1












$begingroup$

Hint. As regards (c) and (e) consider
$$a_n=frac{1+(-1)^n+2^{-n}}{n^2}.$$
Can you modify it in order to obtain a counterexample for (f)?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I apologize, but I do not immediately see how I can modify your example to get a counterexample for (f).
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 7:14






  • 1




    $begingroup$
    @childishsadbino For example replace the numerator of the given $a_n$ with $2+(-1)^n$
    $endgroup$
    – Robert Z
    Dec 11 '18 at 7:17












  • $begingroup$
    Makes sense now! Thank you so much!
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 7:21



















1












$begingroup$

For (c), consider "merging" two convergent series:
$$
a_{2n} = frac 1{3^n}, a_{2n-1}= frac 1{2^n},
$$

then
$$
varlimsup b_n = +infty, varliminf b_n = 0.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This should serve as a counter-example for (e) as well, right?
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 7:02










  • $begingroup$
    Yeah.${{{{{{}}}}}}$
    $endgroup$
    – xbh
    Dec 11 '18 at 7:02












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Hint. As regards (c) and (e) consider
$$a_n=frac{1+(-1)^n+2^{-n}}{n^2}.$$
Can you modify it in order to obtain a counterexample for (f)?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I apologize, but I do not immediately see how I can modify your example to get a counterexample for (f).
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 7:14






  • 1




    $begingroup$
    @childishsadbino For example replace the numerator of the given $a_n$ with $2+(-1)^n$
    $endgroup$
    – Robert Z
    Dec 11 '18 at 7:17












  • $begingroup$
    Makes sense now! Thank you so much!
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 7:21
















1












$begingroup$

Hint. As regards (c) and (e) consider
$$a_n=frac{1+(-1)^n+2^{-n}}{n^2}.$$
Can you modify it in order to obtain a counterexample for (f)?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I apologize, but I do not immediately see how I can modify your example to get a counterexample for (f).
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 7:14






  • 1




    $begingroup$
    @childishsadbino For example replace the numerator of the given $a_n$ with $2+(-1)^n$
    $endgroup$
    – Robert Z
    Dec 11 '18 at 7:17












  • $begingroup$
    Makes sense now! Thank you so much!
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 7:21














1












1








1





$begingroup$

Hint. As regards (c) and (e) consider
$$a_n=frac{1+(-1)^n+2^{-n}}{n^2}.$$
Can you modify it in order to obtain a counterexample for (f)?






share|cite|improve this answer











$endgroup$



Hint. As regards (c) and (e) consider
$$a_n=frac{1+(-1)^n+2^{-n}}{n^2}.$$
Can you modify it in order to obtain a counterexample for (f)?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 11 '18 at 6:58

























answered Dec 11 '18 at 6:52









Robert ZRobert Z

101k1070143




101k1070143












  • $begingroup$
    I apologize, but I do not immediately see how I can modify your example to get a counterexample for (f).
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 7:14






  • 1




    $begingroup$
    @childishsadbino For example replace the numerator of the given $a_n$ with $2+(-1)^n$
    $endgroup$
    – Robert Z
    Dec 11 '18 at 7:17












  • $begingroup$
    Makes sense now! Thank you so much!
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 7:21


















  • $begingroup$
    I apologize, but I do not immediately see how I can modify your example to get a counterexample for (f).
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 7:14






  • 1




    $begingroup$
    @childishsadbino For example replace the numerator of the given $a_n$ with $2+(-1)^n$
    $endgroup$
    – Robert Z
    Dec 11 '18 at 7:17












  • $begingroup$
    Makes sense now! Thank you so much!
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 7:21
















$begingroup$
I apologize, but I do not immediately see how I can modify your example to get a counterexample for (f).
$endgroup$
– childishsadbino
Dec 11 '18 at 7:14




$begingroup$
I apologize, but I do not immediately see how I can modify your example to get a counterexample for (f).
$endgroup$
– childishsadbino
Dec 11 '18 at 7:14




1




1




$begingroup$
@childishsadbino For example replace the numerator of the given $a_n$ with $2+(-1)^n$
$endgroup$
– Robert Z
Dec 11 '18 at 7:17






$begingroup$
@childishsadbino For example replace the numerator of the given $a_n$ with $2+(-1)^n$
$endgroup$
– Robert Z
Dec 11 '18 at 7:17














$begingroup$
Makes sense now! Thank you so much!
$endgroup$
– childishsadbino
Dec 11 '18 at 7:21




$begingroup$
Makes sense now! Thank you so much!
$endgroup$
– childishsadbino
Dec 11 '18 at 7:21











1












$begingroup$

For (c), consider "merging" two convergent series:
$$
a_{2n} = frac 1{3^n}, a_{2n-1}= frac 1{2^n},
$$

then
$$
varlimsup b_n = +infty, varliminf b_n = 0.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This should serve as a counter-example for (e) as well, right?
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 7:02










  • $begingroup$
    Yeah.${{{{{{}}}}}}$
    $endgroup$
    – xbh
    Dec 11 '18 at 7:02
















1












$begingroup$

For (c), consider "merging" two convergent series:
$$
a_{2n} = frac 1{3^n}, a_{2n-1}= frac 1{2^n},
$$

then
$$
varlimsup b_n = +infty, varliminf b_n = 0.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This should serve as a counter-example for (e) as well, right?
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 7:02










  • $begingroup$
    Yeah.${{{{{{}}}}}}$
    $endgroup$
    – xbh
    Dec 11 '18 at 7:02














1












1








1





$begingroup$

For (c), consider "merging" two convergent series:
$$
a_{2n} = frac 1{3^n}, a_{2n-1}= frac 1{2^n},
$$

then
$$
varlimsup b_n = +infty, varliminf b_n = 0.
$$






share|cite|improve this answer









$endgroup$



For (c), consider "merging" two convergent series:
$$
a_{2n} = frac 1{3^n}, a_{2n-1}= frac 1{2^n},
$$

then
$$
varlimsup b_n = +infty, varliminf b_n = 0.
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 11 '18 at 6:59









xbhxbh

6,3201522




6,3201522












  • $begingroup$
    This should serve as a counter-example for (e) as well, right?
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 7:02










  • $begingroup$
    Yeah.${{{{{{}}}}}}$
    $endgroup$
    – xbh
    Dec 11 '18 at 7:02


















  • $begingroup$
    This should serve as a counter-example for (e) as well, right?
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 7:02










  • $begingroup$
    Yeah.${{{{{{}}}}}}$
    $endgroup$
    – xbh
    Dec 11 '18 at 7:02
















$begingroup$
This should serve as a counter-example for (e) as well, right?
$endgroup$
– childishsadbino
Dec 11 '18 at 7:02




$begingroup$
This should serve as a counter-example for (e) as well, right?
$endgroup$
– childishsadbino
Dec 11 '18 at 7:02












$begingroup$
Yeah.${{{{{{}}}}}}$
$endgroup$
– xbh
Dec 11 '18 at 7:02




$begingroup$
Yeah.${{{{{{}}}}}}$
$endgroup$
– xbh
Dec 11 '18 at 7:02


















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