Use comparison test to show $sumfrac{3^n+7^n}{3^n+8^n}$ converges. [closed]
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I want to manipulate it such that $frac{3^n+7^n}{3^n+8^n}leq x^n$ for some $x<1$
sequences-and-series
closed as off-topic by Jyrki Lahtonen, Rebellos, user10354138, Cesareo, Shailesh Nov 20 at 1:12
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I want to manipulate it such that $frac{3^n+7^n}{3^n+8^n}leq x^n$ for some $x<1$
sequences-and-series
closed as off-topic by Jyrki Lahtonen, Rebellos, user10354138, Cesareo, Shailesh Nov 20 at 1:12
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." – Jyrki Lahtonen, Rebellos, user10354138, Cesareo, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
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I want to manipulate it such that $frac{3^n+7^n}{3^n+8^n}leq x^n$ for some $x<1$
sequences-and-series
I want to manipulate it such that $frac{3^n+7^n}{3^n+8^n}leq x^n$ for some $x<1$
sequences-and-series
sequences-and-series
asked Nov 18 at 21:55
m.bazza
827
827
closed as off-topic by Jyrki Lahtonen, Rebellos, user10354138, Cesareo, Shailesh Nov 20 at 1:12
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." – Jyrki Lahtonen, Rebellos, user10354138, Cesareo, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Jyrki Lahtonen, Rebellos, user10354138, Cesareo, Shailesh Nov 20 at 1:12
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." – Jyrki Lahtonen, Rebellos, user10354138, Cesareo, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
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3 Answers
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Notice
$$ frac{ 3^n + 7^n }{3^n + 8^n} leqfrac{3^n+7^n}{8^n} = left( frac{3}{8} right)^n + left( frac{7}{8} right)^n $$
Should I say more?
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We have that eventually
$$frac{3^n+7^n}{3^n+8^n}leq frac{(7.5)^n}{8^n}=left(frac{7.5}{8}right)^n$$
as an alternative by limit comparison test with $sum frac{7^n}{8^n}$ indeed
$$frac{frac{3^n+7^n}{3^n+8^n}}{frac{7^n}{8^n}}=frac{8^n(3^n+7^n)}{7^n(3^n+8^n)} to 1$$
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It is often much easier and quicker to show $|a_n| leq Ccdot x^n$ with a positive constant $C$.
In your case:
$$color{blue}{frac{3^n+7^n}{3^n+8^n}} = left( frac{7}{8} right)^nfrac{left( frac{3}{7} right)^n + 1}{left( frac{3}{8} right)^n + 1} color{blue}{<} left( frac{7}{8} right)^n cdot frac{frac{1}{2} + 1}{1} =color{blue}{frac{3}{2} cdot left( frac{7}{8} right)^n}$$
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Notice
$$ frac{ 3^n + 7^n }{3^n + 8^n} leqfrac{3^n+7^n}{8^n} = left( frac{3}{8} right)^n + left( frac{7}{8} right)^n $$
Should I say more?
add a comment |
up vote
3
down vote
Notice
$$ frac{ 3^n + 7^n }{3^n + 8^n} leqfrac{3^n+7^n}{8^n} = left( frac{3}{8} right)^n + left( frac{7}{8} right)^n $$
Should I say more?
add a comment |
up vote
3
down vote
up vote
3
down vote
Notice
$$ frac{ 3^n + 7^n }{3^n + 8^n} leqfrac{3^n+7^n}{8^n} = left( frac{3}{8} right)^n + left( frac{7}{8} right)^n $$
Should I say more?
Notice
$$ frac{ 3^n + 7^n }{3^n + 8^n} leqfrac{3^n+7^n}{8^n} = left( frac{3}{8} right)^n + left( frac{7}{8} right)^n $$
Should I say more?
answered Nov 18 at 21:59
Jimmy Sabater
1,800218
1,800218
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We have that eventually
$$frac{3^n+7^n}{3^n+8^n}leq frac{(7.5)^n}{8^n}=left(frac{7.5}{8}right)^n$$
as an alternative by limit comparison test with $sum frac{7^n}{8^n}$ indeed
$$frac{frac{3^n+7^n}{3^n+8^n}}{frac{7^n}{8^n}}=frac{8^n(3^n+7^n)}{7^n(3^n+8^n)} to 1$$
add a comment |
up vote
0
down vote
We have that eventually
$$frac{3^n+7^n}{3^n+8^n}leq frac{(7.5)^n}{8^n}=left(frac{7.5}{8}right)^n$$
as an alternative by limit comparison test with $sum frac{7^n}{8^n}$ indeed
$$frac{frac{3^n+7^n}{3^n+8^n}}{frac{7^n}{8^n}}=frac{8^n(3^n+7^n)}{7^n(3^n+8^n)} to 1$$
add a comment |
up vote
0
down vote
up vote
0
down vote
We have that eventually
$$frac{3^n+7^n}{3^n+8^n}leq frac{(7.5)^n}{8^n}=left(frac{7.5}{8}right)^n$$
as an alternative by limit comparison test with $sum frac{7^n}{8^n}$ indeed
$$frac{frac{3^n+7^n}{3^n+8^n}}{frac{7^n}{8^n}}=frac{8^n(3^n+7^n)}{7^n(3^n+8^n)} to 1$$
We have that eventually
$$frac{3^n+7^n}{3^n+8^n}leq frac{(7.5)^n}{8^n}=left(frac{7.5}{8}right)^n$$
as an alternative by limit comparison test with $sum frac{7^n}{8^n}$ indeed
$$frac{frac{3^n+7^n}{3^n+8^n}}{frac{7^n}{8^n}}=frac{8^n(3^n+7^n)}{7^n(3^n+8^n)} to 1$$
answered Nov 18 at 22:04
gimusi
89.7k74495
89.7k74495
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It is often much easier and quicker to show $|a_n| leq Ccdot x^n$ with a positive constant $C$.
In your case:
$$color{blue}{frac{3^n+7^n}{3^n+8^n}} = left( frac{7}{8} right)^nfrac{left( frac{3}{7} right)^n + 1}{left( frac{3}{8} right)^n + 1} color{blue}{<} left( frac{7}{8} right)^n cdot frac{frac{1}{2} + 1}{1} =color{blue}{frac{3}{2} cdot left( frac{7}{8} right)^n}$$
add a comment |
up vote
0
down vote
It is often much easier and quicker to show $|a_n| leq Ccdot x^n$ with a positive constant $C$.
In your case:
$$color{blue}{frac{3^n+7^n}{3^n+8^n}} = left( frac{7}{8} right)^nfrac{left( frac{3}{7} right)^n + 1}{left( frac{3}{8} right)^n + 1} color{blue}{<} left( frac{7}{8} right)^n cdot frac{frac{1}{2} + 1}{1} =color{blue}{frac{3}{2} cdot left( frac{7}{8} right)^n}$$
add a comment |
up vote
0
down vote
up vote
0
down vote
It is often much easier and quicker to show $|a_n| leq Ccdot x^n$ with a positive constant $C$.
In your case:
$$color{blue}{frac{3^n+7^n}{3^n+8^n}} = left( frac{7}{8} right)^nfrac{left( frac{3}{7} right)^n + 1}{left( frac{3}{8} right)^n + 1} color{blue}{<} left( frac{7}{8} right)^n cdot frac{frac{1}{2} + 1}{1} =color{blue}{frac{3}{2} cdot left( frac{7}{8} right)^n}$$
It is often much easier and quicker to show $|a_n| leq Ccdot x^n$ with a positive constant $C$.
In your case:
$$color{blue}{frac{3^n+7^n}{3^n+8^n}} = left( frac{7}{8} right)^nfrac{left( frac{3}{7} right)^n + 1}{left( frac{3}{8} right)^n + 1} color{blue}{<} left( frac{7}{8} right)^n cdot frac{frac{1}{2} + 1}{1} =color{blue}{frac{3}{2} cdot left( frac{7}{8} right)^n}$$
answered Nov 19 at 8:43
trancelocation
8,5821520
8,5821520
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