Mixing
convergence of series : rational with exponentials [closed]
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I have this infinite series below, not sure how to handle the alternating term in the bottom. I am guessing i could use triangle inequality on the bottom, then the absolute of the alternating term would become positive.
Not sure if i can replace or remove this term. Need some help
$$
sum frac7^n(-2)^n +5^n
$$
calculus sequences-and-series
edited Jul 27 at 17:19
mvw
30.2k22250
asked Jul 27 at 17:12
Palu
2952620
closed as off-topic by user21820, Did, Mostafa Ayaz, Strants, José Carlos Santos Jul 30 at 22:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." Рuser21820, Did, Mostafa Ayaz, Strants, Jos̩ Carlos Santos
add a comment |Â
up vote
1
down vote
favorite
I have this infinite series below, not sure how to handle the alternating term in the bottom. I am guessing i could use triangle inequality on the bottom, then the absolute of the alternating term would become positive.
Not sure if i can replace or remove this term. Need some help
$$
sum frac7^n(-2)^n +5^n
$$
calculus sequences-and-series
edited Jul 27 at 17:19
mvw
30.2k22250
asked Jul 27 at 17:12
Palu
2952620
closed as off-topic by user21820, Did, Mostafa Ayaz, Strants, José Carlos Santos Jul 30 at 22:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." Рuser21820, Did, Mostafa Ayaz, Strants, Jos̩ Carlos Santos
2
What is the question?
– José Carlos Santos
Jul 27 at 17:13
Hint: think about the behavior of your terms for large $n$.
– lulu
Jul 27 at 17:14
I might've had a little brain fart on my answer lol
– Rushabh Mehta
Jul 27 at 17:23
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have this infinite series below, not sure how to handle the alternating term in the bottom. I am guessing i could use triangle inequality on the bottom, then the absolute of the alternating term would become positive.
Not sure if i can replace or remove this term. Need some help
$$
sum frac7^n(-2)^n +5^n
$$
calculus sequences-and-series
edited Jul 27 at 17:19
mvw
30.2k22250
asked Jul 27 at 17:12
Palu
2952620
I have this infinite series below, not sure how to handle the alternating term in the bottom. I am guessing i could use triangle inequality on the bottom, then the absolute of the alternating term would become positive.
Not sure if i can replace or remove this term. Need some help
$$
sum frac7^n(-2)^n +5^n
$$
calculus sequences-and-series
edited Jul 27 at 17:19
mvw
30.2k22250
asked Jul 27 at 17:12
Palu
2952620
edited Jul 27 at 17:19
mvw
30.2k22250
edited Jul 27 at 17:19
mvw
30.2k22250
edited Jul 27 at 17:19
mvw
30.2k22250
30.2k22250
asked Jul 27 at 17:12
Palu
2952620
asked Jul 27 at 17:12
Palu
2952620
asked Jul 27 at 17:12
Palu
2952620
2952620
closed as off-topic by user21820, Did, Mostafa Ayaz, Strants, José Carlos Santos Jul 30 at 22:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." Рuser21820, Did, Mostafa Ayaz, Strants, Jos̩ Carlos Santos
closed as off-topic by user21820, Did, Mostafa Ayaz, Strants, José Carlos Santos Jul 30 at 22:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." Рuser21820, Did, Mostafa Ayaz, Strants, Jos̩ Carlos Santos
2
What is the question?
– José Carlos Santos
Jul 27 at 17:13
Hint: think about the behavior of your terms for large $n$.
– lulu
Jul 27 at 17:14
I might've had a little brain fart on my answer lol
– Rushabh Mehta
Jul 27 at 17:23
add a comment |Â
2
What is the question?
– José Carlos Santos
Jul 27 at 17:13
Hint: think about the behavior of your terms for large $n$.
– lulu
Jul 27 at 17:14
I might've had a little brain fart on my answer lol
– Rushabh Mehta
Jul 27 at 17:23
2
2
What is the question?
– José Carlos Santos
Jul 27 at 17:13
What is the question?
– José Carlos Santos
Jul 27 at 17:13
Hint: think about the behavior of your terms for large $n$.
– lulu
Jul 27 at 17:14
Hint: think about the behavior of your terms for large $n$.
– lulu
Jul 27 at 17:14
I might've had a little brain fart on my answer lol
– Rushabh Mehta
Jul 27 at 17:23
I might've had a little brain fart on my answer lol
– Rushabh Mehta
Jul 27 at 17:23
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
The simplest way would be to notice that
$$
5^n + (-2)^n = 5^n + (-1)^ncdot 2^n leq 5^n + 2^n leq 5^n + 5^n = 2cdot 5^n
$$
for every $ngeq 1$, and therefore
$$
sum_n=1^N frac7^n5^n + (-2)^n geq sum_n=1^N frac7^n2cdot 5^n = frac12sum_n=1^N left(frac75right)^n,.$$
Now, for $Ntoinfty$, the RHS diverges to $infty$, so...
answered Jul 27 at 17:19
Clement C.
47k33682
Hi Clement, I like the way you did the inequality work on this and taking out the (-1)^n.
– Palu
Jul 27 at 17:40
Thanks. Glad that helped!
– Clement C.
Jul 27 at 17:47
Hi Clement, since we constructed here a minimum bound of the original function, does that imply that if the minimum bound diverges then the actual series diverges.
– Palu
Jul 28 at 4:46
OK, i see that is the case, i looked up the comparison test in detail, and see that is the case.
– Palu
Jul 28 at 4:48
Yes (this is a simple case of the squeeze theorem, if you will). If $a_n geq b_n$ and $b_ntoinfty$, then $a_n to infty$.
– Clement C.
Jul 28 at 4:48
 |Â
show 1 more comment
up vote
1
down vote
It suffices to observe that
$$a_n=frac7^n(-2)^n +5^nge frac7^n2^n +5^n=frac(2+5)^n2^n +5^nge1$$
and therefore $$sum_n=1^N a_nge N to infty$$
answered Jul 27 at 17:25
gimusi
64.9k73483
Hi, I am not sure why some people or a moderator put this on hold. I asked a question and people were able to answer this OK. And I chose the answer i thought was the best. SO I am surprised that people would put on HOLD a question that is already answered. I really don't see any issues at all. This question seems fairly clear. SO very odd that my questions are put on HOLD when they are already answered. I can understand if there is no answer being awarded and then it being put on hold.
– Palu
Aug 2 at 18:19
@Palu I agree and voted for reopening. Bye
– gimusi
Aug 2 at 18:27
Thanks for your support gimusi. Really appreciate your input here.
– Palu
Aug 2 at 18:32
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The simplest way would be to notice that
$$
5^n + (-2)^n = 5^n + (-1)^ncdot 2^n leq 5^n + 2^n leq 5^n + 5^n = 2cdot 5^n
$$
for every $ngeq 1$, and therefore
$$
sum_n=1^N frac7^n5^n + (-2)^n geq sum_n=1^N frac7^n2cdot 5^n = frac12sum_n=1^N left(frac75right)^n,.$$
Now, for $Ntoinfty$, the RHS diverges to $infty$, so...
answered Jul 27 at 17:19
Clement C.
47k33682
Hi Clement, I like the way you did the inequality work on this and taking out the (-1)^n.
– Palu
Jul 27 at 17:40
Thanks. Glad that helped!
– Clement C.
Jul 27 at 17:47
Hi Clement, since we constructed here a minimum bound of the original function, does that imply that if the minimum bound diverges then the actual series diverges.
– Palu
Jul 28 at 4:46
OK, i see that is the case, i looked up the comparison test in detail, and see that is the case.
– Palu
Jul 28 at 4:48
Yes (this is a simple case of the squeeze theorem, if you will). If $a_n geq b_n$ and $b_ntoinfty$, then $a_n to infty$.
– Clement C.
Jul 28 at 4:48
 |Â
show 1 more comment
up vote
2
down vote
accepted
The simplest way would be to notice that
$$
5^n + (-2)^n = 5^n + (-1)^ncdot 2^n leq 5^n + 2^n leq 5^n + 5^n = 2cdot 5^n
$$
for every $ngeq 1$, and therefore
$$
sum_n=1^N frac7^n5^n + (-2)^n geq sum_n=1^N frac7^n2cdot 5^n = frac12sum_n=1^N left(frac75right)^n,.$$
Now, for $Ntoinfty$, the RHS diverges to $infty$, so...
answered Jul 27 at 17:19
Clement C.
47k33682
Hi Clement, I like the way you did the inequality work on this and taking out the (-1)^n.
– Palu
Jul 27 at 17:40
Thanks. Glad that helped!
– Clement C.
Jul 27 at 17:47
Hi Clement, since we constructed here a minimum bound of the original function, does that imply that if the minimum bound diverges then the actual series diverges.
– Palu
Jul 28 at 4:46
OK, i see that is the case, i looked up the comparison test in detail, and see that is the case.
– Palu
Jul 28 at 4:48
Yes (this is a simple case of the squeeze theorem, if you will). If $a_n geq b_n$ and $b_ntoinfty$, then $a_n to infty$.
– Clement C.
Jul 28 at 4:48
 |Â
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The simplest way would be to notice that
$$
5^n + (-2)^n = 5^n + (-1)^ncdot 2^n leq 5^n + 2^n leq 5^n + 5^n = 2cdot 5^n
$$
for every $ngeq 1$, and therefore
$$
sum_n=1^N frac7^n5^n + (-2)^n geq sum_n=1^N frac7^n2cdot 5^n = frac12sum_n=1^N left(frac75right)^n,.$$
Now, for $Ntoinfty$, the RHS diverges to $infty$, so...
answered Jul 27 at 17:19
Clement C.
47k33682
The simplest way would be to notice that
$$
5^n + (-2)^n = 5^n + (-1)^ncdot 2^n leq 5^n + 2^n leq 5^n + 5^n = 2cdot 5^n
$$
for every $ngeq 1$, and therefore
$$
sum_n=1^N frac7^n5^n + (-2)^n geq sum_n=1^N frac7^n2cdot 5^n = frac12sum_n=1^N left(frac75right)^n,.$$
Now, for $Ntoinfty$, the RHS diverges to $infty$, so...
answered Jul 27 at 17:19
Clement C.
47k33682
answered Jul 27 at 17:19
Clement C.
47k33682
answered Jul 27 at 17:19
Clement C.
47k33682
answered Jul 27 at 17:19
Clement C.
47k33682
47k33682
Hi Clement, I like the way you did the inequality work on this and taking out the (-1)^n.
– Palu
Jul 27 at 17:40
Thanks. Glad that helped!
– Clement C.
Jul 27 at 17:47
Hi Clement, since we constructed here a minimum bound of the original function, does that imply that if the minimum bound diverges then the actual series diverges.
– Palu
Jul 28 at 4:46
OK, i see that is the case, i looked up the comparison test in detail, and see that is the case.
– Palu
Jul 28 at 4:48
Yes (this is a simple case of the squeeze theorem, if you will). If $a_n geq b_n$ and $b_ntoinfty$, then $a_n to infty$.
– Clement C.
Jul 28 at 4:48
 |Â
show 1 more comment
Hi Clement, I like the way you did the inequality work on this and taking out the (-1)^n.
– Palu
Jul 27 at 17:40
Thanks. Glad that helped!
– Clement C.
Jul 27 at 17:47
Hi Clement, since we constructed here a minimum bound of the original function, does that imply that if the minimum bound diverges then the actual series diverges.
– Palu
Jul 28 at 4:46
OK, i see that is the case, i looked up the comparison test in detail, and see that is the case.
– Palu
Jul 28 at 4:48
Yes (this is a simple case of the squeeze theorem, if you will). If $a_n geq b_n$ and $b_ntoinfty$, then $a_n to infty$.
– Clement C.
Jul 28 at 4:48
Hi Clement, I like the way you did the inequality work on this and taking out the (-1)^n.
– Palu
Jul 27 at 17:40
Hi Clement, I like the way you did the inequality work on this and taking out the (-1)^n.
– Palu
Jul 27 at 17:40
Thanks. Glad that helped!
– Clement C.
Jul 27 at 17:47
Thanks. Glad that helped!
– Clement C.
Jul 27 at 17:47
Hi Clement, since we constructed here a minimum bound of the original function, does that imply that if the minimum bound diverges then the actual series diverges.
– Palu
Jul 28 at 4:46
Hi Clement, since we constructed here a minimum bound of the original function, does that imply that if the minimum bound diverges then the actual series diverges.
– Palu
Jul 28 at 4:46
OK, i see that is the case, i looked up the comparison test in detail, and see that is the case.
– Palu
Jul 28 at 4:48
OK, i see that is the case, i looked up the comparison test in detail, and see that is the case.
– Palu
Jul 28 at 4:48
Yes (this is a simple case of the squeeze theorem, if you will). If $a_n geq b_n$ and $b_ntoinfty$, then $a_n to infty$.
– Clement C.
Jul 28 at 4:48
Yes (this is a simple case of the squeeze theorem, if you will). If $a_n geq b_n$ and $b_ntoinfty$, then $a_n to infty$.
– Clement C.
Jul 28 at 4:48
 |Â
show 1 more comment
up vote
1
down vote
It suffices to observe that
$$a_n=frac7^n(-2)^n +5^nge frac7^n2^n +5^n=frac(2+5)^n2^n +5^nge1$$
and therefore $$sum_n=1^N a_nge N to infty$$
answered Jul 27 at 17:25
gimusi
64.9k73483
Hi, I am not sure why some people or a moderator put this on hold. I asked a question and people were able to answer this OK. And I chose the answer i thought was the best. SO I am surprised that people would put on HOLD a question that is already answered. I really don't see any issues at all. This question seems fairly clear. SO very odd that my questions are put on HOLD when they are already answered. I can understand if there is no answer being awarded and then it being put on hold.
– Palu
Aug 2 at 18:19
@Palu I agree and voted for reopening. Bye
– gimusi
Aug 2 at 18:27
Thanks for your support gimusi. Really appreciate your input here.
– Palu
Aug 2 at 18:32
add a comment |Â
up vote
1
down vote
It suffices to observe that
$$a_n=frac7^n(-2)^n +5^nge frac7^n2^n +5^n=frac(2+5)^n2^n +5^nge1$$
and therefore $$sum_n=1^N a_nge N to infty$$
answered Jul 27 at 17:25
gimusi
64.9k73483
Hi, I am not sure why some people or a moderator put this on hold. I asked a question and people were able to answer this OK. And I chose the answer i thought was the best. SO I am surprised that people would put on HOLD a question that is already answered. I really don't see any issues at all. This question seems fairly clear. SO very odd that my questions are put on HOLD when they are already answered. I can understand if there is no answer being awarded and then it being put on hold.
– Palu
Aug 2 at 18:19
@Palu I agree and voted for reopening. Bye
– gimusi
Aug 2 at 18:27
Thanks for your support gimusi. Really appreciate your input here.
– Palu
Aug 2 at 18:32
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It suffices to observe that
$$a_n=frac7^n(-2)^n +5^nge frac7^n2^n +5^n=frac(2+5)^n2^n +5^nge1$$
and therefore $$sum_n=1^N a_nge N to infty$$
answered Jul 27 at 17:25
gimusi
64.9k73483
It suffices to observe that
$$a_n=frac7^n(-2)^n +5^nge frac7^n2^n +5^n=frac(2+5)^n2^n +5^nge1$$
and therefore $$sum_n=1^N a_nge N to infty$$
answered Jul 27 at 17:25
gimusi
64.9k73483
edited Jul 27 at 17:30
answered Jul 27 at 17:25
gimusi
64.9k73483
answered Jul 27 at 17:25
gimusi
64.9k73483
answered Jul 27 at 17:25
gimusi
64.9k73483
64.9k73483
Hi, I am not sure why some people or a moderator put this on hold. I asked a question and people were able to answer this OK. And I chose the answer i thought was the best. SO I am surprised that people would put on HOLD a question that is already answered. I really don't see any issues at all. This question seems fairly clear. SO very odd that my questions are put on HOLD when they are already answered. I can understand if there is no answer being awarded and then it being put on hold.
– Palu
Aug 2 at 18:19
@Palu I agree and voted for reopening. Bye
– gimusi
Aug 2 at 18:27
Thanks for your support gimusi. Really appreciate your input here.
– Palu
Aug 2 at 18:32
add a comment |Â
Hi, I am not sure why some people or a moderator put this on hold. I asked a question and people were able to answer this OK. And I chose the answer i thought was the best. SO I am surprised that people would put on HOLD a question that is already answered. I really don't see any issues at all. This question seems fairly clear. SO very odd that my questions are put on HOLD when they are already answered. I can understand if there is no answer being awarded and then it being put on hold.
– Palu
Aug 2 at 18:19
@Palu I agree and voted for reopening. Bye
– gimusi
Aug 2 at 18:27
Thanks for your support gimusi. Really appreciate your input here.
– Palu
Aug 2 at 18:32
Hi, I am not sure why some people or a moderator put this on hold. I asked a question and people were able to answer this OK. And I chose the answer i thought was the best. SO I am surprised that people would put on HOLD a question that is already answered. I really don't see any issues at all. This question seems fairly clear. SO very odd that my questions are put on HOLD when they are already answered. I can understand if there is no answer being awarded and then it being put on hold.
– Palu
Aug 2 at 18:19
Hi, I am not sure why some people or a moderator put this on hold. I asked a question and people were able to answer this OK. And I chose the answer i thought was the best. SO I am surprised that people would put on HOLD a question that is already answered. I really don't see any issues at all. This question seems fairly clear. SO very odd that my questions are put on HOLD when they are already answered. I can understand if there is no answer being awarded and then it being put on hold.
– Palu
Aug 2 at 18:19
@Palu I agree and voted for reopening. Bye
– gimusi
Aug 2 at 18:27
@Palu I agree and voted for reopening. Bye
– gimusi
Aug 2 at 18:27
Thanks for your support gimusi. Really appreciate your input here.
– Palu
Aug 2 at 18:32
Thanks for your support gimusi. Really appreciate your input here.
– Palu
Aug 2 at 18:32
add a comment |Â
2
What is the question?
– José Carlos Santos
Jul 27 at 17:13
Hint: think about the behavior of your terms for large $n$.
– lulu
Jul 27 at 17:14
I might've had a little brain fart on my answer lol
– Rushabh Mehta
Jul 27 at 17:23