Finding if $sum_{n=1}^inftyfrac{n^n+1/n}{(n+1/n)^n}$ is convergent or divergent .
1
Is this series convergent or divergent ?
(The sum will be infinite or it will converge to a certain number?)
$$sum_{n=1}^inftyfrac{n^n+1/n}{(n+1/n)^n}$$
calculus sequences-and-series
edited Nov 28 '18 at 15:45
Asaf Karagila♦
302k32426757
asked Nov 28 '18 at 14:36
NegarNegar
397
add a comment |
1
Is this series convergent or divergent ?
(The sum will be infinite or it will converge to a certain number?)
$$sum_{n=1}^inftyfrac{n^n+1/n}{(n+1/n)^n}$$
calculus sequences-and-series
edited Nov 28 '18 at 15:45
Asaf Karagila♦
302k32426757
asked Nov 28 '18 at 14:36
NegarNegar
397
Welcome to MathStackExchange! Please take a look at the tour page and consider editing your question to comply with the homework questions guidelines. In particular, it is good practice to give some context on what course you are taking and show your work in attempting to solve this question.
– Mefitico
Nov 28 '18 at 14:43
add a comment |
1
1
1
Is this series convergent or divergent ?
(The sum will be infinite or it will converge to a certain number?)
$$sum_{n=1}^inftyfrac{n^n+1/n}{(n+1/n)^n}$$
calculus sequences-and-series
edited Nov 28 '18 at 15:45
Asaf Karagila♦
302k32426757
asked Nov 28 '18 at 14:36
NegarNegar
397
Is this series convergent or divergent ?
(The sum will be infinite or it will converge to a certain number?)
$$sum_{n=1}^inftyfrac{n^n+1/n}{(n+1/n)^n}$$
calculus sequences-and-series
calculus sequences-and-series
edited Nov 28 '18 at 15:45
Asaf Karagila♦
302k32426757
asked Nov 28 '18 at 14:36
NegarNegar
397
edited Nov 28 '18 at 15:45
Asaf Karagila♦
302k32426757
asked Nov 28 '18 at 14:36
NegarNegar
397
edited Nov 28 '18 at 15:45
Asaf Karagila♦
302k32426757
edited Nov 28 '18 at 15:45
Asaf Karagila♦
302k32426757
edited Nov 28 '18 at 15:45
Asaf Karagila♦
302k32426757
302k32426757
asked Nov 28 '18 at 14:36
NegarNegar
397
asked Nov 28 '18 at 14:36
NegarNegar
397
asked Nov 28 '18 at 14:36
NegarNegar
397
397
Welcome to MathStackExchange! Please take a look at the tour page and consider editing your question to comply with the homework questions guidelines. In particular, it is good practice to give some context on what course you are taking and show your work in attempting to solve this question.
– Mefitico
Nov 28 '18 at 14:43
add a comment |
Welcome to MathStackExchange! Please take a look at the tour page and consider editing your question to comply with the homework questions guidelines. In particular, it is good practice to give some context on what course you are taking and show your work in attempting to solve this question.
– Mefitico
Nov 28 '18 at 14:43
Welcome to MathStackExchange! Please take a look at the tour page and consider editing your question to comply with the homework questions guidelines. In particular, it is good practice to give some context on what course you are taking and show your work in attempting to solve this question.
– Mefitico
Nov 28 '18 at 14:43
Welcome to MathStackExchange! Please take a look at the tour page and consider editing your question to comply with the homework questions guidelines. In particular, it is good practice to give some context on what course you are taking and show your work in attempting to solve this question.
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2 Answers
2
active
oldest
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4
You can certainly ignore the term $dfrac1n$ at the numerator and
$$sumfrac{n^n}{(n+frac1n)^n}=sumfrac1{(1+frac1{n^2})^n}$$ diverges as the general term tends to $1$.
answered Nov 28 '18 at 14:42
Yves DaoustYves Daoust
124k671222
1
At first, I lost the n^n term at the numerator.
– gimusi
Nov 28 '18 at 14:52
Shouldn’t we use one of the series tests to answer this question?
– Negar
Nov 28 '18 at 15:12
@Negarrezaei: it is enough to note that as $(1+1/n^2)^n=sqrt[n]{(1+1/n^2)^{n^2}}$, the general term goes to $1$.
– Yves Daoust
Nov 28 '18 at 15:15
@YvesDaoust even befor simplifying , the sum of the very first 3 characters of the series would be 2.45. How can the sum of the whole characters of the series from 1 to infinitive would be convergent to 1 ?
– Negar
Nov 28 '18 at 15:39
@Negar: I am talking about the general term, not the partial sums. And I said that the sum diverges !!!
– Yves Daoust
Nov 28 '18 at 16:00
|
show 1 more comment
-1
HINT
We have that
$$(n+1/n)^n=e^{nlog (n+1/n)}=e^{nlog n+nlog(1+1/n^2)}sim n^n$$
$$frac{n^n+1/n}{(n+1/n)^n}sim1$$
answered Nov 28 '18 at 14:42
gimusigimusi
1
We have sum in the question.I don’t get how the sum of the whole characters of the series is converge to 1 when even the sum of the very fisrt 3 characters would be 2,45 ?
– Negar
Nov 28 '18 at 15:57
@Negar We are claiming that for n large the terms of the series is equal to $1$ and then the series diverges.
– gimusi
Nov 28 '18 at 15:59
I got that thanks.
– Negar
Nov 28 '18 at 16:02
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
4
You can certainly ignore the term $dfrac1n$ at the numerator and
$$sumfrac{n^n}{(n+frac1n)^n}=sumfrac1{(1+frac1{n^2})^n}$$ diverges as the general term tends to $1$.
answered Nov 28 '18 at 14:42
Yves DaoustYves Daoust
124k671222
1
At first, I lost the n^n term at the numerator.
– gimusi
Nov 28 '18 at 14:52
Shouldn’t we use one of the series tests to answer this question?
– Negar
Nov 28 '18 at 15:12
@Negarrezaei: it is enough to note that as $(1+1/n^2)^n=sqrt[n]{(1+1/n^2)^{n^2}}$, the general term goes to $1$.
– Yves Daoust
Nov 28 '18 at 15:15
@YvesDaoust even befor simplifying , the sum of the very first 3 characters of the series would be 2.45. How can the sum of the whole characters of the series from 1 to infinitive would be convergent to 1 ?
– Negar
Nov 28 '18 at 15:39
@Negar: I am talking about the general term, not the partial sums. And I said that the sum diverges !!!
– Yves Daoust
Nov 28 '18 at 16:00
|
show 1 more comment
4
You can certainly ignore the term $dfrac1n$ at the numerator and
$$sumfrac{n^n}{(n+frac1n)^n}=sumfrac1{(1+frac1{n^2})^n}$$ diverges as the general term tends to $1$.
answered Nov 28 '18 at 14:42
Yves DaoustYves Daoust
124k671222
1
At first, I lost the n^n term at the numerator.
– gimusi
Nov 28 '18 at 14:52
Shouldn’t we use one of the series tests to answer this question?
– Negar
Nov 28 '18 at 15:12
@Negarrezaei: it is enough to note that as $(1+1/n^2)^n=sqrt[n]{(1+1/n^2)^{n^2}}$, the general term goes to $1$.
– Yves Daoust
Nov 28 '18 at 15:15
@YvesDaoust even befor simplifying , the sum of the very first 3 characters of the series would be 2.45. How can the sum of the whole characters of the series from 1 to infinitive would be convergent to 1 ?
– Negar
Nov 28 '18 at 15:39
@Negar: I am talking about the general term, not the partial sums. And I said that the sum diverges !!!
– Yves Daoust
Nov 28 '18 at 16:00
|
show 1 more comment
4
4
4
You can certainly ignore the term $dfrac1n$ at the numerator and
$$sumfrac{n^n}{(n+frac1n)^n}=sumfrac1{(1+frac1{n^2})^n}$$ diverges as the general term tends to $1$.
answered Nov 28 '18 at 14:42
Yves DaoustYves Daoust
124k671222
You can certainly ignore the term $dfrac1n$ at the numerator and
$$sumfrac{n^n}{(n+frac1n)^n}=sumfrac1{(1+frac1{n^2})^n}$$ diverges as the general term tends to $1$.
answered Nov 28 '18 at 14:42
Yves DaoustYves Daoust
124k671222
answered Nov 28 '18 at 14:42
Yves DaoustYves Daoust
124k671222
answered Nov 28 '18 at 14:42
Yves DaoustYves Daoust
124k671222
answered Nov 28 '18 at 14:42
Yves DaoustYves Daoust
124k671222
124k671222
1
At first, I lost the n^n term at the numerator.
– gimusi
Nov 28 '18 at 14:52
Shouldn’t we use one of the series tests to answer this question?
– Negar
Nov 28 '18 at 15:12
@Negarrezaei: it is enough to note that as $(1+1/n^2)^n=sqrt[n]{(1+1/n^2)^{n^2}}$, the general term goes to $1$.
– Yves Daoust
Nov 28 '18 at 15:15
@YvesDaoust even befor simplifying , the sum of the very first 3 characters of the series would be 2.45. How can the sum of the whole characters of the series from 1 to infinitive would be convergent to 1 ?
– Negar
Nov 28 '18 at 15:39
@Negar: I am talking about the general term, not the partial sums. And I said that the sum diverges !!!
– Yves Daoust
Nov 28 '18 at 16:00
|
show 1 more comment
1
At first, I lost the n^n term at the numerator.
– gimusi
Nov 28 '18 at 14:52
Shouldn’t we use one of the series tests to answer this question?
– Negar
Nov 28 '18 at 15:12
@Negarrezaei: it is enough to note that as $(1+1/n^2)^n=sqrt[n]{(1+1/n^2)^{n^2}}$, the general term goes to $1$.
– Yves Daoust
Nov 28 '18 at 15:15
@YvesDaoust even befor simplifying , the sum of the very first 3 characters of the series would be 2.45. How can the sum of the whole characters of the series from 1 to infinitive would be convergent to 1 ?
– Negar
Nov 28 '18 at 15:39
@Negar: I am talking about the general term, not the partial sums. And I said that the sum diverges !!!
– Yves Daoust
Nov 28 '18 at 16:00
1
1
At first, I lost the n^n term at the numerator.
– gimusi
Nov 28 '18 at 14:52
At first, I lost the n^n term at the numerator.
– gimusi
Nov 28 '18 at 14:52
Shouldn’t we use one of the series tests to answer this question?
– Negar
Nov 28 '18 at 15:12
Shouldn’t we use one of the series tests to answer this question?
– Negar
Nov 28 '18 at 15:12
@Negarrezaei: it is enough to note that as $(1+1/n^2)^n=sqrt[n]{(1+1/n^2)^{n^2}}$, the general term goes to $1$.
– Yves Daoust
Nov 28 '18 at 15:15
@Negarrezaei: it is enough to note that as $(1+1/n^2)^n=sqrt[n]{(1+1/n^2)^{n^2}}$, the general term goes to $1$.
– Yves Daoust
Nov 28 '18 at 15:15
@YvesDaoust even befor simplifying , the sum of the very first 3 characters of the series would be 2.45. How can the sum of the whole characters of the series from 1 to infinitive would be convergent to 1 ?
– Negar
Nov 28 '18 at 15:39
@YvesDaoust even befor simplifying , the sum of the very first 3 characters of the series would be 2.45. How can the sum of the whole characters of the series from 1 to infinitive would be convergent to 1 ?
– Negar
Nov 28 '18 at 15:39
@Negar: I am talking about the general term, not the partial sums. And I said that the sum diverges !!!
– Yves Daoust
Nov 28 '18 at 16:00
@Negar: I am talking about the general term, not the partial sums. And I said that the sum diverges !!!
– Yves Daoust
Nov 28 '18 at 16:00
|
show 1 more comment
-1
HINT
We have that
$$(n+1/n)^n=e^{nlog (n+1/n)}=e^{nlog n+nlog(1+1/n^2)}sim n^n$$
$$frac{n^n+1/n}{(n+1/n)^n}sim1$$
answered Nov 28 '18 at 14:42
gimusigimusi
1
We have sum in the question.I don’t get how the sum of the whole characters of the series is converge to 1 when even the sum of the very fisrt 3 characters would be 2,45 ?
– Negar
Nov 28 '18 at 15:57
@Negar We are claiming that for n large the terms of the series is equal to $1$ and then the series diverges.
– gimusi
Nov 28 '18 at 15:59
I got that thanks.
– Negar
Nov 28 '18 at 16:02
add a comment |
-1
HINT
We have that
$$(n+1/n)^n=e^{nlog (n+1/n)}=e^{nlog n+nlog(1+1/n^2)}sim n^n$$
$$frac{n^n+1/n}{(n+1/n)^n}sim1$$
answered Nov 28 '18 at 14:42
gimusigimusi
1
We have sum in the question.I don’t get how the sum of the whole characters of the series is converge to 1 when even the sum of the very fisrt 3 characters would be 2,45 ?
– Negar
Nov 28 '18 at 15:57
@Negar We are claiming that for n large the terms of the series is equal to $1$ and then the series diverges.
– gimusi
Nov 28 '18 at 15:59
I got that thanks.
– Negar
Nov 28 '18 at 16:02
add a comment |
-1
-1
-1
HINT
We have that
$$(n+1/n)^n=e^{nlog (n+1/n)}=e^{nlog n+nlog(1+1/n^2)}sim n^n$$
$$frac{n^n+1/n}{(n+1/n)^n}sim1$$
answered Nov 28 '18 at 14:42
gimusigimusi
1
HINT
We have that
$$(n+1/n)^n=e^{nlog (n+1/n)}=e^{nlog n+nlog(1+1/n^2)}sim n^n$$
$$frac{n^n+1/n}{(n+1/n)^n}sim1$$
answered Nov 28 '18 at 14:42
gimusigimusi
1
edited Nov 28 '18 at 14:48
answered Nov 28 '18 at 14:42
gimusigimusi
1
answered Nov 28 '18 at 14:42
gimusigimusi
1
answered Nov 28 '18 at 14:42
gimusigimusi
1
1
We have sum in the question.I don’t get how the sum of the whole characters of the series is converge to 1 when even the sum of the very fisrt 3 characters would be 2,45 ?
– Negar
Nov 28 '18 at 15:57
@Negar We are claiming that for n large the terms of the series is equal to $1$ and then the series diverges.
– gimusi
Nov 28 '18 at 15:59
I got that thanks.
– Negar
Nov 28 '18 at 16:02
add a comment |
We have sum in the question.I don’t get how the sum of the whole characters of the series is converge to 1 when even the sum of the very fisrt 3 characters would be 2,45 ?
– Negar
Nov 28 '18 at 15:57
@Negar We are claiming that for n large the terms of the series is equal to $1$ and then the series diverges.
– gimusi
Nov 28 '18 at 15:59
I got that thanks.
– Negar
Nov 28 '18 at 16:02
We have sum in the question.I don’t get how the sum of the whole characters of the series is converge to 1 when even the sum of the very fisrt 3 characters would be 2,45 ?
– Negar
Nov 28 '18 at 15:57
We have sum in the question.I don’t get how the sum of the whole characters of the series is converge to 1 when even the sum of the very fisrt 3 characters would be 2,45 ?
– Negar
Nov 28 '18 at 15:57
@Negar We are claiming that for n large the terms of the series is equal to $1$ and then the series diverges.
– gimusi
Nov 28 '18 at 15:59
@Negar We are claiming that for n large the terms of the series is equal to $1$ and then the series diverges.
– gimusi
Nov 28 '18 at 15:59
I got that thanks.
– Negar
Nov 28 '18 at 16:02
I got that thanks.
– Negar
Nov 28 '18 at 16:02
add a comment |
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Welcome to MathStackExchange! Please take a look at the tour page and consider editing your question to comply with the homework questions guidelines. In particular, it is good practice to give some context on what course you are taking and show your work in attempting to solve this question.
– Mefitico
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