Convergent Series: $sqrtn+1/(n^2 + 1)$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Series convergence divergence and tests and sumsDetermine whether the series $sum _n=1^infty :fracn^2-5nsqrtn^7+2n+1$ is convergent or divergent.Alternating series: $sumlimits_n= 1^infty (-1)^n-1 fracln(n)n$ convergence?How to figure out if this series is convergent or divergent?Infinite series, convergent or divergent.Cal $2$ $p$-series convergent vs divergentRatio test when checking the convergence of seriesSequences and series and the alternating series testSeries $sum n!exp(-n^2)$ is convergent?Sequences and Series: Find the value for x such that series (ln x)^n is convergent
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Convergent Series: $sqrtn+1/(n^2 + 1)$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Series convergence divergence and tests and sumsDetermine whether the series $sum _n=1^infty :fracn^2-5nsqrtn^7+2n+1$ is convergent or divergent.Alternating series: $sumlimits_n= 1^infty (-1)^n-1 fracln(n)n$ convergence?How to figure out if this series is convergent or divergent?Infinite series, convergent or divergent.Cal $2$ $p$-series convergent vs divergentRatio test when checking the convergence of seriesSequences and series and the alternating series testSeries $sum n!exp(-n^2)$ is convergent?Sequences and Series: Find the value for x such that series (ln x)^n is convergent
$begingroup$
I have this homework problem that I'm having difficulty on.
$$
sum_n=1^infty fracsqrtn+1n^2+2
$$
I've found that the series is most likely convergent by the divergent test. As using lhospitals rule, the sequence approaches 0, thus it MIGHT be a convergent series. However, I do think it's convergent because it doesn't appear to be a harmonic sequence. So now I need to find the sum of the series, and that's where I am having trouble with.
sequences-and-series convergence
edited Mar 25 at 19:15
José Carlos Santos
175k23134243
asked Mar 25 at 19:01
Christian MartinezChristian Martinez
607
$endgroup$
add a comment |
$begingroup$
I have this homework problem that I'm having difficulty on.
$$
sum_n=1^infty fracsqrtn+1n^2+2
$$
I've found that the series is most likely convergent by the divergent test. As using lhospitals rule, the sequence approaches 0, thus it MIGHT be a convergent series. However, I do think it's convergent because it doesn't appear to be a harmonic sequence. So now I need to find the sum of the series, and that's where I am having trouble with.
sequences-and-series convergence
edited Mar 25 at 19:15
José Carlos Santos
175k23134243
asked Mar 25 at 19:01
Christian MartinezChristian Martinez
607
$endgroup$
1
$begingroup$
The series is in fact convergent, you just need to compare the general term with $1/n^alpha$, with convenient $alpha$. This is a very basic exercise, you should try to do it yourself before asking for the solution in a forum, otherwise you will not learn much from this.
$endgroup$
– PierreCarre
Mar 25 at 19:06
1
$begingroup$
Welcome to Math Stack Exchange. I noticed the title has $n^2+1$ but the question has $n^2+2$
$endgroup$
– J. W. Tanner
Mar 25 at 20:03
add a comment |
$begingroup$
I have this homework problem that I'm having difficulty on.
$$
sum_n=1^infty fracsqrtn+1n^2+2
$$
I've found that the series is most likely convergent by the divergent test. As using lhospitals rule, the sequence approaches 0, thus it MIGHT be a convergent series. However, I do think it's convergent because it doesn't appear to be a harmonic sequence. So now I need to find the sum of the series, and that's where I am having trouble with.
sequences-and-series convergence
edited Mar 25 at 19:15
José Carlos Santos
175k23134243
asked Mar 25 at 19:01
Christian MartinezChristian Martinez
607
$endgroup$
I have this homework problem that I'm having difficulty on.
$$
sum_n=1^infty fracsqrtn+1n^2+2
$$
I've found that the series is most likely convergent by the divergent test. As using lhospitals rule, the sequence approaches 0, thus it MIGHT be a convergent series. However, I do think it's convergent because it doesn't appear to be a harmonic sequence. So now I need to find the sum of the series, and that's where I am having trouble with.
sequences-and-series convergence
sequences-and-series convergence
edited Mar 25 at 19:15
José Carlos Santos
175k23134243
asked Mar 25 at 19:01
Christian MartinezChristian Martinez
607
edited Mar 25 at 19:15
José Carlos Santos
175k23134243
asked Mar 25 at 19:01
Christian MartinezChristian Martinez
607
edited Mar 25 at 19:15
José Carlos Santos
175k23134243
edited Mar 25 at 19:15
José Carlos Santos
175k23134243
edited Mar 25 at 19:15
José Carlos Santos
175k23134243
175k23134243
asked Mar 25 at 19:01
Christian MartinezChristian Martinez
607
asked Mar 25 at 19:01
Christian MartinezChristian Martinez
607
asked Mar 25 at 19:01
Christian MartinezChristian Martinez
607
607
1
$begingroup$
The series is in fact convergent, you just need to compare the general term with $1/n^alpha$, with convenient $alpha$. This is a very basic exercise, you should try to do it yourself before asking for the solution in a forum, otherwise you will not learn much from this.
$endgroup$
– PierreCarre
Mar 25 at 19:06
1
$begingroup$
Welcome to Math Stack Exchange. I noticed the title has $n^2+1$ but the question has $n^2+2$
$endgroup$
– J. W. Tanner
Mar 25 at 20:03
add a comment |
1
$begingroup$
The series is in fact convergent, you just need to compare the general term with $1/n^alpha$, with convenient $alpha$. This is a very basic exercise, you should try to do it yourself before asking for the solution in a forum, otherwise you will not learn much from this.
$endgroup$
– PierreCarre
Mar 25 at 19:06
1
$begingroup$
Welcome to Math Stack Exchange. I noticed the title has $n^2+1$ but the question has $n^2+2$
$endgroup$
– J. W. Tanner
Mar 25 at 20:03
1
1
$begingroup$
The series is in fact convergent, you just need to compare the general term with $1/n^alpha$, with convenient $alpha$. This is a very basic exercise, you should try to do it yourself before asking for the solution in a forum, otherwise you will not learn much from this.
$endgroup$
– PierreCarre
Mar 25 at 19:06
$begingroup$
The series is in fact convergent, you just need to compare the general term with $1/n^alpha$, with convenient $alpha$. This is a very basic exercise, you should try to do it yourself before asking for the solution in a forum, otherwise you will not learn much from this.
$endgroup$
– PierreCarre
Mar 25 at 19:06
1
1
$begingroup$
Welcome to Math Stack Exchange. I noticed the title has $n^2+1$ but the question has $n^2+2$
$endgroup$
– J. W. Tanner
Mar 25 at 20:03
$begingroup$
Welcome to Math Stack Exchange. I noticed the title has $n^2+1$ but the question has $n^2+2$
$endgroup$
– J. W. Tanner
Mar 25 at 20:03
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: Compute the limit$$lim_ntoinftyfracdfracsqrtn+1n^2+1dfrac1n^3/2.$$
answered Mar 25 at 19:04
José Carlos SantosJosé Carlos Santos
175k23134243
$endgroup$
$begingroup$
I'm not sure how you got there, could you please elaborate?
$endgroup$
– Christian Martinez
Mar 25 at 19:24
$begingroup$
That limit is a real number greater than $0$. Therefore, either both series are convergent or they both diverge.
$endgroup$
– José Carlos Santos
Mar 25 at 19:27
add a comment |
$begingroup$
Note that for $nge1$, we have the estimate
$$0<fracsqrtn+1n^2+2le fracsqrtn+nn^2=fracsqrt2n^3/2 $$
Inasmuch as the series $sum_nge 1frac1n^3/2$ converges, the series of interest, $sum_nge1 fracsqrtn+1n^2+2$ does likewise.
answered Mar 25 at 21:49
Mark ViolaMark Viola
134k1278177
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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votes
$begingroup$
Hint: Compute the limit$$lim_ntoinftyfracdfracsqrtn+1n^2+1dfrac1n^3/2.$$
answered Mar 25 at 19:04
José Carlos SantosJosé Carlos Santos
175k23134243
$endgroup$
$begingroup$
I'm not sure how you got there, could you please elaborate?
$endgroup$
– Christian Martinez
Mar 25 at 19:24
$begingroup$
That limit is a real number greater than $0$. Therefore, either both series are convergent or they both diverge.
$endgroup$
– José Carlos Santos
Mar 25 at 19:27
add a comment |
$begingroup$
Hint: Compute the limit$$lim_ntoinftyfracdfracsqrtn+1n^2+1dfrac1n^3/2.$$
answered Mar 25 at 19:04
José Carlos SantosJosé Carlos Santos
175k23134243
$endgroup$
$begingroup$
I'm not sure how you got there, could you please elaborate?
$endgroup$
– Christian Martinez
Mar 25 at 19:24
$begingroup$
That limit is a real number greater than $0$. Therefore, either both series are convergent or they both diverge.
$endgroup$
– José Carlos Santos
Mar 25 at 19:27
add a comment |
$begingroup$
Hint: Compute the limit$$lim_ntoinftyfracdfracsqrtn+1n^2+1dfrac1n^3/2.$$
answered Mar 25 at 19:04
José Carlos SantosJosé Carlos Santos
175k23134243
$endgroup$
Hint: Compute the limit$$lim_ntoinftyfracdfracsqrtn+1n^2+1dfrac1n^3/2.$$
answered Mar 25 at 19:04
José Carlos SantosJosé Carlos Santos
175k23134243
answered Mar 25 at 19:04
José Carlos SantosJosé Carlos Santos
175k23134243
answered Mar 25 at 19:04
José Carlos SantosJosé Carlos Santos
175k23134243
answered Mar 25 at 19:04
José Carlos SantosJosé Carlos Santos
175k23134243
175k23134243
$begingroup$
I'm not sure how you got there, could you please elaborate?
$endgroup$
– Christian Martinez
Mar 25 at 19:24
$begingroup$
That limit is a real number greater than $0$. Therefore, either both series are convergent or they both diverge.
$endgroup$
– José Carlos Santos
Mar 25 at 19:27
add a comment |
$begingroup$
I'm not sure how you got there, could you please elaborate?
$endgroup$
– Christian Martinez
Mar 25 at 19:24
$begingroup$
That limit is a real number greater than $0$. Therefore, either both series are convergent or they both diverge.
$endgroup$
– José Carlos Santos
Mar 25 at 19:27
$begingroup$
I'm not sure how you got there, could you please elaborate?
$endgroup$
– Christian Martinez
Mar 25 at 19:24
$begingroup$
I'm not sure how you got there, could you please elaborate?
$endgroup$
– Christian Martinez
Mar 25 at 19:24
$begingroup$
That limit is a real number greater than $0$. Therefore, either both series are convergent or they both diverge.
$endgroup$
– José Carlos Santos
Mar 25 at 19:27
$begingroup$
That limit is a real number greater than $0$. Therefore, either both series are convergent or they both diverge.
$endgroup$
– José Carlos Santos
Mar 25 at 19:27
add a comment |
$begingroup$
Note that for $nge1$, we have the estimate
$$0<fracsqrtn+1n^2+2le fracsqrtn+nn^2=fracsqrt2n^3/2 $$
Inasmuch as the series $sum_nge 1frac1n^3/2$ converges, the series of interest, $sum_nge1 fracsqrtn+1n^2+2$ does likewise.
answered Mar 25 at 21:49
Mark ViolaMark Viola
134k1278177
$endgroup$
add a comment |
$begingroup$
Note that for $nge1$, we have the estimate
$$0<fracsqrtn+1n^2+2le fracsqrtn+nn^2=fracsqrt2n^3/2 $$
Inasmuch as the series $sum_nge 1frac1n^3/2$ converges, the series of interest, $sum_nge1 fracsqrtn+1n^2+2$ does likewise.
answered Mar 25 at 21:49
Mark ViolaMark Viola
134k1278177
$endgroup$
add a comment |
$begingroup$
Note that for $nge1$, we have the estimate
$$0<fracsqrtn+1n^2+2le fracsqrtn+nn^2=fracsqrt2n^3/2 $$
Inasmuch as the series $sum_nge 1frac1n^3/2$ converges, the series of interest, $sum_nge1 fracsqrtn+1n^2+2$ does likewise.
answered Mar 25 at 21:49
Mark ViolaMark Viola
134k1278177
$endgroup$
Note that for $nge1$, we have the estimate
$$0<fracsqrtn+1n^2+2le fracsqrtn+nn^2=fracsqrt2n^3/2 $$
Inasmuch as the series $sum_nge 1frac1n^3/2$ converges, the series of interest, $sum_nge1 fracsqrtn+1n^2+2$ does likewise.
answered Mar 25 at 21:49
Mark ViolaMark Viola
134k1278177
answered Mar 25 at 21:49
Mark ViolaMark Viola
134k1278177
answered Mar 25 at 21:49
Mark ViolaMark Viola
134k1278177
answered Mar 25 at 21:49
Mark ViolaMark Viola
134k1278177
134k1278177
add a comment |
add a comment |
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$begingroup$
The series is in fact convergent, you just need to compare the general term with $1/n^alpha$, with convenient $alpha$. This is a very basic exercise, you should try to do it yourself before asking for the solution in a forum, otherwise you will not learn much from this.
$endgroup$
– PierreCarre
Mar 25 at 19:06
1
$begingroup$
Welcome to Math Stack Exchange. I noticed the title has $n^2+1$ but the question has $n^2+2$
$endgroup$
– J. W. Tanner
Mar 25 at 20:03