Mixing
Applying the integral test for the series $sum _{n =1}^{infty} frac{1}{n ln n}$
$begingroup$
Applying the integral test for the series $sum _{n =1}^{infty} frac{1}{n ln n}$, I am wondering what will I do when $n = 1$ I think I will get an improper integral in both the integrand and the upper limit of the integration, then what shall I do, could anyone tell methe details of this please?
real-analysis sequences-and-series
edited Jan 29 at 14:46
YuiTo Cheng
2,1862937
asked Jan 29 at 7:14
hopefullyhopefully
299214
$endgroup$
add a comment |
$begingroup$
Applying the integral test for the series $sum _{n =1}^{infty} frac{1}{n ln n}$, I am wondering what will I do when $n = 1$ I think I will get an improper integral in both the integrand and the upper limit of the integration, then what shall I do, could anyone tell methe details of this please?
real-analysis sequences-and-series
edited Jan 29 at 14:46
YuiTo Cheng
2,1862937
asked Jan 29 at 7:14
hopefullyhopefully
299214
$endgroup$
1
$begingroup$
Start the sum at $n=2$.
$endgroup$
– Lord Shark the Unknown
Jan 29 at 7:14
$begingroup$
so the first term does not matter with me .... correct?@ChinnapparajR
$endgroup$
– hopefully
Jan 29 at 7:17
1
$begingroup$
The sum itself is not defined for the $n=1$ term. If this was presented in some textbook, it is probably just an oversight, and you should consider the sum starting at $n=2$.
$endgroup$
– alex.jordan
Jan 29 at 7:17
add a comment |
$begingroup$
Applying the integral test for the series $sum _{n =1}^{infty} frac{1}{n ln n}$, I am wondering what will I do when $n = 1$ I think I will get an improper integral in both the integrand and the upper limit of the integration, then what shall I do, could anyone tell methe details of this please?
real-analysis sequences-and-series
edited Jan 29 at 14:46
YuiTo Cheng
2,1862937
asked Jan 29 at 7:14
hopefullyhopefully
299214
$endgroup$
Applying the integral test for the series $sum _{n =1}^{infty} frac{1}{n ln n}$, I am wondering what will I do when $n = 1$ I think I will get an improper integral in both the integrand and the upper limit of the integration, then what shall I do, could anyone tell methe details of this please?
real-analysis sequences-and-series
real-analysis sequences-and-series
edited Jan 29 at 14:46
YuiTo Cheng
2,1862937
asked Jan 29 at 7:14
hopefullyhopefully
299214
edited Jan 29 at 14:46
YuiTo Cheng
2,1862937
asked Jan 29 at 7:14
hopefullyhopefully
299214
edited Jan 29 at 14:46
YuiTo Cheng
2,1862937
edited Jan 29 at 14:46
YuiTo Cheng
2,1862937
edited Jan 29 at 14:46
YuiTo Cheng
2,1862937
2,1862937
asked Jan 29 at 7:14
hopefullyhopefully
299214
asked Jan 29 at 7:14
hopefullyhopefully
299214
asked Jan 29 at 7:14
hopefullyhopefully
299214
299214
1
$begingroup$
Start the sum at $n=2$.
$endgroup$
– Lord Shark the Unknown
Jan 29 at 7:14
$begingroup$
so the first term does not matter with me .... correct?@ChinnapparajR
$endgroup$
– hopefully
Jan 29 at 7:17
1
$begingroup$
The sum itself is not defined for the $n=1$ term. If this was presented in some textbook, it is probably just an oversight, and you should consider the sum starting at $n=2$.
$endgroup$
– alex.jordan
Jan 29 at 7:17
add a comment |
1
$begingroup$
Start the sum at $n=2$.
$endgroup$
– Lord Shark the Unknown
Jan 29 at 7:14
$begingroup$
so the first term does not matter with me .... correct?@ChinnapparajR
$endgroup$
– hopefully
Jan 29 at 7:17
1
$begingroup$
The sum itself is not defined for the $n=1$ term. If this was presented in some textbook, it is probably just an oversight, and you should consider the sum starting at $n=2$.
$endgroup$
– alex.jordan
Jan 29 at 7:17
1
1
$begingroup$
Start the sum at $n=2$.
$endgroup$
– Lord Shark the Unknown
Jan 29 at 7:14
$begingroup$
Start the sum at $n=2$.
$endgroup$
– Lord Shark the Unknown
Jan 29 at 7:14
$begingroup$
so the first term does not matter with me .... correct?@ChinnapparajR
$endgroup$
– hopefully
Jan 29 at 7:17
$begingroup$
so the first term does not matter with me .... correct?@ChinnapparajR
$endgroup$
– hopefully
Jan 29 at 7:17
1
1
$begingroup$
The sum itself is not defined for the $n=1$ term. If this was presented in some textbook, it is probably just an oversight, and you should consider the sum starting at $n=2$.
$endgroup$
– alex.jordan
Jan 29 at 7:17
$begingroup$
The sum itself is not defined for the $n=1$ term. If this was presented in some textbook, it is probably just an oversight, and you should consider the sum starting at $n=2$.
$endgroup$
– alex.jordan
Jan 29 at 7:17
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The general term is not defined for $n=1$, so that the sum as is presented is not defined. However, it is well defined if you start at $n=2$.
Anyway, if you had any other problem when applying some test for
$$sum_{n=2}^infty a_n$$
(for instance, if you had to compute $int_2^infty f(x) dx$ for some $f$ with discontinuities in the interval $(2,infty)$, or if some necessary hypothesis only were valid from some point on), remember that this series will converge if and only if
$$sum_{n=k}^infty a_n,$$
for any $kge 2$. This means you can always choose to omit an initial interval as needed.
For instance, you can't apply the integral test to the series
$$sum_{n=0}^infty frac1{sqrt{|n^3-2|}}$$
by computing
$$int_0^infty frac1{sqrt{|x^3-2|}}dx$$
(give it a thought to this, there might be more than one reason),
but you can apply the test to
$$sum_{n=2}^infty frac1{sqrt{|n^3-2|}}$$
by computing
$$int_2^infty frac1{sqrt{|x^3-2|}}dx,$$
and using the fact that both series will converge or both won't (in this case, they're both well defined).
answered Jan 29 at 7:54
Alejandro Nasif SalumAlejandro Nasif Salum
4,775118
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The general term is not defined for $n=1$, so that the sum as is presented is not defined. However, it is well defined if you start at $n=2$.
Anyway, if you had any other problem when applying some test for
$$sum_{n=2}^infty a_n$$
(for instance, if you had to compute $int_2^infty f(x) dx$ for some $f$ with discontinuities in the interval $(2,infty)$, or if some necessary hypothesis only were valid from some point on), remember that this series will converge if and only if
$$sum_{n=k}^infty a_n,$$
for any $kge 2$. This means you can always choose to omit an initial interval as needed.
For instance, you can't apply the integral test to the series
$$sum_{n=0}^infty frac1{sqrt{|n^3-2|}}$$
by computing
$$int_0^infty frac1{sqrt{|x^3-2|}}dx$$
(give it a thought to this, there might be more than one reason),
but you can apply the test to
$$sum_{n=2}^infty frac1{sqrt{|n^3-2|}}$$
by computing
$$int_2^infty frac1{sqrt{|x^3-2|}}dx,$$
and using the fact that both series will converge or both won't (in this case, they're both well defined).
answered Jan 29 at 7:54
Alejandro Nasif SalumAlejandro Nasif Salum
4,775118
$endgroup$
add a comment |
$begingroup$
The general term is not defined for $n=1$, so that the sum as is presented is not defined. However, it is well defined if you start at $n=2$.
Anyway, if you had any other problem when applying some test for
$$sum_{n=2}^infty a_n$$
(for instance, if you had to compute $int_2^infty f(x) dx$ for some $f$ with discontinuities in the interval $(2,infty)$, or if some necessary hypothesis only were valid from some point on), remember that this series will converge if and only if
$$sum_{n=k}^infty a_n,$$
for any $kge 2$. This means you can always choose to omit an initial interval as needed.
For instance, you can't apply the integral test to the series
$$sum_{n=0}^infty frac1{sqrt{|n^3-2|}}$$
by computing
$$int_0^infty frac1{sqrt{|x^3-2|}}dx$$
(give it a thought to this, there might be more than one reason),
but you can apply the test to
$$sum_{n=2}^infty frac1{sqrt{|n^3-2|}}$$
by computing
$$int_2^infty frac1{sqrt{|x^3-2|}}dx,$$
and using the fact that both series will converge or both won't (in this case, they're both well defined).
answered Jan 29 at 7:54
Alejandro Nasif SalumAlejandro Nasif Salum
4,775118
$endgroup$
add a comment |
$begingroup$
The general term is not defined for $n=1$, so that the sum as is presented is not defined. However, it is well defined if you start at $n=2$.
Anyway, if you had any other problem when applying some test for
$$sum_{n=2}^infty a_n$$
(for instance, if you had to compute $int_2^infty f(x) dx$ for some $f$ with discontinuities in the interval $(2,infty)$, or if some necessary hypothesis only were valid from some point on), remember that this series will converge if and only if
$$sum_{n=k}^infty a_n,$$
for any $kge 2$. This means you can always choose to omit an initial interval as needed.
For instance, you can't apply the integral test to the series
$$sum_{n=0}^infty frac1{sqrt{|n^3-2|}}$$
by computing
$$int_0^infty frac1{sqrt{|x^3-2|}}dx$$
(give it a thought to this, there might be more than one reason),
but you can apply the test to
$$sum_{n=2}^infty frac1{sqrt{|n^3-2|}}$$
by computing
$$int_2^infty frac1{sqrt{|x^3-2|}}dx,$$
and using the fact that both series will converge or both won't (in this case, they're both well defined).
answered Jan 29 at 7:54
Alejandro Nasif SalumAlejandro Nasif Salum
4,775118
$endgroup$
The general term is not defined for $n=1$, so that the sum as is presented is not defined. However, it is well defined if you start at $n=2$.
Anyway, if you had any other problem when applying some test for
$$sum_{n=2}^infty a_n$$
(for instance, if you had to compute $int_2^infty f(x) dx$ for some $f$ with discontinuities in the interval $(2,infty)$, or if some necessary hypothesis only were valid from some point on), remember that this series will converge if and only if
$$sum_{n=k}^infty a_n,$$
for any $kge 2$. This means you can always choose to omit an initial interval as needed.
For instance, you can't apply the integral test to the series
$$sum_{n=0}^infty frac1{sqrt{|n^3-2|}}$$
by computing
$$int_0^infty frac1{sqrt{|x^3-2|}}dx$$
(give it a thought to this, there might be more than one reason),
but you can apply the test to
$$sum_{n=2}^infty frac1{sqrt{|n^3-2|}}$$
by computing
$$int_2^infty frac1{sqrt{|x^3-2|}}dx,$$
and using the fact that both series will converge or both won't (in this case, they're both well defined).
answered Jan 29 at 7:54
Alejandro Nasif SalumAlejandro Nasif Salum
4,775118
answered Jan 29 at 7:54
Alejandro Nasif SalumAlejandro Nasif Salum
4,775118
answered Jan 29 at 7:54
Alejandro Nasif SalumAlejandro Nasif Salum
4,775118
answered Jan 29 at 7:54
Alejandro Nasif SalumAlejandro Nasif Salum
4,775118
4,775118
add a comment |
add a comment |
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1
$begingroup$
Start the sum at $n=2$.
$endgroup$
– Lord Shark the Unknown
Jan 29 at 7:14
$begingroup$
so the first term does not matter with me .... correct?@ChinnapparajR
$endgroup$
– hopefully
Jan 29 at 7:17
1
$begingroup$
The sum itself is not defined for the $n=1$ term. If this was presented in some textbook, it is probably just an oversight, and you should consider the sum starting at $n=2$.
$endgroup$
– alex.jordan
Jan 29 at 7:17