Prove $sum a_n b_n$ diverges if $a_n$ diverges, $a_n>0$, and $liminf_n b_n >0$
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I'm having trouble starting this proof. From the initial hypothesis, we know for some $n>N$, $b_n>0$. Since $a_n>0$ and $a_n$ diverges, for some $n>N'$, $a_ngeq varepsilon$. Then $a_n b_n>0$, so if $sum a_n b_n$ is a series of positive numbers. I'm not sure how to show it diverges however, so any help would be appreciated!
real-analysis sequences-and-series convergence
asked Dec 3 at 2:51
t.perez
387
add a comment |
up vote
1
down vote
favorite
I'm having trouble starting this proof. From the initial hypothesis, we know for some $n>N$, $b_n>0$. Since $a_n>0$ and $a_n$ diverges, for some $n>N'$, $a_ngeq varepsilon$. Then $a_n b_n>0$, so if $sum a_n b_n$ is a series of positive numbers. I'm not sure how to show it diverges however, so any help would be appreciated!
real-analysis sequences-and-series convergence
asked Dec 3 at 2:51
t.perez
387
@qbert isn't the answer below correct regardless?
– clark
Dec 3 at 3:03
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm having trouble starting this proof. From the initial hypothesis, we know for some $n>N$, $b_n>0$. Since $a_n>0$ and $a_n$ diverges, for some $n>N'$, $a_ngeq varepsilon$. Then $a_n b_n>0$, so if $sum a_n b_n$ is a series of positive numbers. I'm not sure how to show it diverges however, so any help would be appreciated!
real-analysis sequences-and-series convergence
asked Dec 3 at 2:51
t.perez
387
I'm having trouble starting this proof. From the initial hypothesis, we know for some $n>N$, $b_n>0$. Since $a_n>0$ and $a_n$ diverges, for some $n>N'$, $a_ngeq varepsilon$. Then $a_n b_n>0$, so if $sum a_n b_n$ is a series of positive numbers. I'm not sure how to show it diverges however, so any help would be appreciated!
real-analysis sequences-and-series convergence
real-analysis sequences-and-series convergence
asked Dec 3 at 2:51
t.perez
387
asked Dec 3 at 2:51
t.perez
387
asked Dec 3 at 2:51
t.perez
387
asked Dec 3 at 2:51
t.perez
387
asked Dec 3 at 2:51
t.perez
387
387
@qbert isn't the answer below correct regardless?
– clark
Dec 3 at 3:03
add a comment |
@qbert isn't the answer below correct regardless?
– clark
Dec 3 at 3:03
@qbert isn't the answer below correct regardless?
– clark
Dec 3 at 3:03
@qbert isn't the answer below correct regardless?
– clark
Dec 3 at 3:03
add a comment |
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
Let
$$
liminf_{nto infty}b_n=2delta>0
$$
Then, for large enough $N$, $b_ngeq delta$ for all $ngeq N$. Then,
$$
sum_{n=N}^infty b_na_ngeq deltasum_{n=N}^infty a_n=+infty
$$
since $lim_{nto infty}a_nne 0$.
answered Dec 3 at 3:00
qbert
21.9k32459
Thank you! My only question is, where does the 2 go in the $2delta$ term?
– t.perez
Dec 5 at 2:12
okay, that makes sense now. thank you!
– t.perez
Dec 5 at 2:24
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Let
$$
liminf_{nto infty}b_n=2delta>0
$$
Then, for large enough $N$, $b_ngeq delta$ for all $ngeq N$. Then,
$$
sum_{n=N}^infty b_na_ngeq deltasum_{n=N}^infty a_n=+infty
$$
since $lim_{nto infty}a_nne 0$.
answered Dec 3 at 3:00
qbert
21.9k32459
Thank you! My only question is, where does the 2 go in the $2delta$ term?
– t.perez
Dec 5 at 2:12
okay, that makes sense now. thank you!
– t.perez
Dec 5 at 2:24
add a comment |
up vote
4
down vote
accepted
Let
$$
liminf_{nto infty}b_n=2delta>0
$$
Then, for large enough $N$, $b_ngeq delta$ for all $ngeq N$. Then,
$$
sum_{n=N}^infty b_na_ngeq deltasum_{n=N}^infty a_n=+infty
$$
since $lim_{nto infty}a_nne 0$.
answered Dec 3 at 3:00
qbert
21.9k32459
Thank you! My only question is, where does the 2 go in the $2delta$ term?
– t.perez
Dec 5 at 2:12
okay, that makes sense now. thank you!
– t.perez
Dec 5 at 2:24
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Let
$$
liminf_{nto infty}b_n=2delta>0
$$
Then, for large enough $N$, $b_ngeq delta$ for all $ngeq N$. Then,
$$
sum_{n=N}^infty b_na_ngeq deltasum_{n=N}^infty a_n=+infty
$$
since $lim_{nto infty}a_nne 0$.
answered Dec 3 at 3:00
qbert
21.9k32459
Let
$$
liminf_{nto infty}b_n=2delta>0
$$
Then, for large enough $N$, $b_ngeq delta$ for all $ngeq N$. Then,
$$
sum_{n=N}^infty b_na_ngeq deltasum_{n=N}^infty a_n=+infty
$$
since $lim_{nto infty}a_nne 0$.
answered Dec 3 at 3:00
qbert
21.9k32459
edited Dec 3 at 4:08
answered Dec 3 at 3:00
qbert
21.9k32459
answered Dec 3 at 3:00
qbert
21.9k32459
answered Dec 3 at 3:00
qbert
21.9k32459
21.9k32459
Thank you! My only question is, where does the 2 go in the $2delta$ term?
– t.perez
Dec 5 at 2:12
okay, that makes sense now. thank you!
– t.perez
Dec 5 at 2:24
add a comment |
Thank you! My only question is, where does the 2 go in the $2delta$ term?
– t.perez
Dec 5 at 2:12
okay, that makes sense now. thank you!
– t.perez
Dec 5 at 2:24
Thank you! My only question is, where does the 2 go in the $2delta$ term?
– t.perez
Dec 5 at 2:12
Thank you! My only question is, where does the 2 go in the $2delta$ term?
– t.perez
Dec 5 at 2:12
okay, that makes sense now. thank you!
– t.perez
Dec 5 at 2:24
okay, that makes sense now. thank you!
– t.perez
Dec 5 at 2:24
add a comment |
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@qbert isn't the answer below correct regardless?
– clark
Dec 3 at 3:03