Infinite Series $sumlimits_{n=2}^infty frac{sqrt n (n+1)}{(n^2+1)(n-1)}$ [on hold]
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$$sum_{n=2}^infty frac{sqrt n (n+1)}{(n^2+1)(n-1)}$$
I am trying to solve this problem, I am supposed to determine whether the series converges or diverges but i am not sure which method I should do, I was thinking the ratio test but not entirely sure if that would work
calculus sequences-and-series convergence
edited Dec 2 at 4:22
Martin Sleziak
44.5k7115268
asked Dec 1 at 21:37
M.Jordan
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put on hold as off-topic by RRL, Brahadeesh, user10354138, Jyrki Lahtonen, Rebellos Dec 2 at 9:28
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." – RRL, Brahadeesh, user10354138, Jyrki Lahtonen, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
0
down vote
favorite
$$sum_{n=2}^infty frac{sqrt n (n+1)}{(n^2+1)(n-1)}$$
I am trying to solve this problem, I am supposed to determine whether the series converges or diverges but i am not sure which method I should do, I was thinking the ratio test but not entirely sure if that would work
calculus sequences-and-series convergence
edited Dec 2 at 4:22
Martin Sleziak
44.5k7115268
asked Dec 1 at 21:37
M.Jordan
1
New contributor
M.Jordan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by RRL, Brahadeesh, user10354138, Jyrki Lahtonen, Rebellos Dec 2 at 9:28
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." – RRL, Brahadeesh, user10354138, Jyrki Lahtonen, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
Use D'Alembert criterion and compute $lim_{ntoinfty}frac{a_{n+1}}{a_n}$
– Tito Eliatron
Dec 1 at 21:39
2
Welcome to MSE! If possible, you should learn to use MathJax so that you can type your problem into the question rather than giving a link: math.meta.stackexchange.com/questions/5020/…
– Frpzzd
Dec 1 at 21:39
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$$sum_{n=2}^infty frac{sqrt n (n+1)}{(n^2+1)(n-1)}$$
I am trying to solve this problem, I am supposed to determine whether the series converges or diverges but i am not sure which method I should do, I was thinking the ratio test but not entirely sure if that would work
calculus sequences-and-series convergence
edited Dec 2 at 4:22
Martin Sleziak
44.5k7115268
asked Dec 1 at 21:37
M.Jordan
1
New contributor
M.Jordan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$$sum_{n=2}^infty frac{sqrt n (n+1)}{(n^2+1)(n-1)}$$
I am trying to solve this problem, I am supposed to determine whether the series converges or diverges but i am not sure which method I should do, I was thinking the ratio test but not entirely sure if that would work
calculus sequences-and-series convergence
calculus sequences-and-series convergence
edited Dec 2 at 4:22
Martin Sleziak
44.5k7115268
asked Dec 1 at 21:37
M.Jordan
1
New contributor
M.Jordan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 2 at 4:22
Martin Sleziak
44.5k7115268
asked Dec 1 at 21:37
M.Jordan
1
New contributor
M.Jordan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 2 at 4:22
Martin Sleziak
44.5k7115268
edited Dec 2 at 4:22
Martin Sleziak
44.5k7115268
edited Dec 2 at 4:22
Martin Sleziak
44.5k7115268
44.5k7115268
asked Dec 1 at 21:37
M.Jordan
1
New contributor
M.Jordan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 1 at 21:37
M.Jordan
1
asked Dec 1 at 21:37
M.Jordan
1
1
New contributor
M.Jordan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
M.Jordan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
M.Jordan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by RRL, Brahadeesh, user10354138, Jyrki Lahtonen, Rebellos Dec 2 at 9:28
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." – RRL, Brahadeesh, user10354138, Jyrki Lahtonen, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by RRL, Brahadeesh, user10354138, Jyrki Lahtonen, Rebellos Dec 2 at 9:28
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." – RRL, Brahadeesh, user10354138, Jyrki Lahtonen, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
Use D'Alembert criterion and compute $lim_{ntoinfty}frac{a_{n+1}}{a_n}$
– Tito Eliatron
Dec 1 at 21:39
2
Welcome to MSE! If possible, you should learn to use MathJax so that you can type your problem into the question rather than giving a link: math.meta.stackexchange.com/questions/5020/…
– Frpzzd
Dec 1 at 21:39
add a comment |
Use D'Alembert criterion and compute $lim_{ntoinfty}frac{a_{n+1}}{a_n}$
– Tito Eliatron
Dec 1 at 21:39
2
Welcome to MSE! If possible, you should learn to use MathJax so that you can type your problem into the question rather than giving a link: math.meta.stackexchange.com/questions/5020/…
– Frpzzd
Dec 1 at 21:39
Use D'Alembert criterion and compute $lim_{ntoinfty}frac{a_{n+1}}{a_n}$
– Tito Eliatron
Dec 1 at 21:39
Use D'Alembert criterion and compute $lim_{ntoinfty}frac{a_{n+1}}{a_n}$
– Tito Eliatron
Dec 1 at 21:39
2
2
Welcome to MSE! If possible, you should learn to use MathJax so that you can type your problem into the question rather than giving a link: math.meta.stackexchange.com/questions/5020/…
– Frpzzd
Dec 1 at 21:39
Welcome to MSE! If possible, you should learn to use MathJax so that you can type your problem into the question rather than giving a link: math.meta.stackexchange.com/questions/5020/…
– Frpzzd
Dec 1 at 21:39
add a comment |
2 Answers
2
active
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votes
up vote
1
down vote
HINT
We have that
$$frac{sqrt n (n+1)}{(n^2+1)(n-1)} sim frac{n^frac32}{n^3}=frac1{n^frac32}$$
then refer to limit comparison test.
answered Dec 1 at 21:39
gimusi
90.3k74495
thanks ill give it a try
– M.Jordan
Dec 1 at 21:42
@M.Jordan What about $sum frac1{n^frac32}$? Does it converge or diverge?
– gimusi
Dec 1 at 21:43
that would be a p series so it would coverge since its greater than 1 ?
– M.Jordan
Dec 1 at 21:44
Yes exactly $sum frac1{n^p}$ converges for any $p>1$ and the given series behaves as that for $n$ large. To formalize that in a proof by LCT we need to show that $$lim_{nto infty} frac{frac{sqrt n (n+1)}{(n^2+1)(n-1)}}{frac1{n^frac32}}=lim_{nto infty}frac{n^frac32sqrt n (n+1)}{(n^2+1)(n-1)}=Lin mathbb{R}$$
– gimusi
Dec 1 at 21:47
add a comment |
up vote
1
down vote
We have
$$
frac{n+1}{n-1}=frac{n-1+2}{n-1}=1+frac{2}{n-1}le 3
$$
Also
$$
frac{1}{n^2+1}<frac{1}{n^2}
$$
Hence, for $nge2$,
$$
0<frac{sqrt{n}(n+1)}{(n^2+1)(n-1)}<frac{3}{n^{3/2}}
$$
Can you finish?
answered Dec 1 at 22:05
egreg
175k1383198
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
HINT
We have that
$$frac{sqrt n (n+1)}{(n^2+1)(n-1)} sim frac{n^frac32}{n^3}=frac1{n^frac32}$$
then refer to limit comparison test.
answered Dec 1 at 21:39
gimusi
90.3k74495
thanks ill give it a try
– M.Jordan
Dec 1 at 21:42
@M.Jordan What about $sum frac1{n^frac32}$? Does it converge or diverge?
– gimusi
Dec 1 at 21:43
that would be a p series so it would coverge since its greater than 1 ?
– M.Jordan
Dec 1 at 21:44
Yes exactly $sum frac1{n^p}$ converges for any $p>1$ and the given series behaves as that for $n$ large. To formalize that in a proof by LCT we need to show that $$lim_{nto infty} frac{frac{sqrt n (n+1)}{(n^2+1)(n-1)}}{frac1{n^frac32}}=lim_{nto infty}frac{n^frac32sqrt n (n+1)}{(n^2+1)(n-1)}=Lin mathbb{R}$$
– gimusi
Dec 1 at 21:47
add a comment |
up vote
1
down vote
HINT
We have that
$$frac{sqrt n (n+1)}{(n^2+1)(n-1)} sim frac{n^frac32}{n^3}=frac1{n^frac32}$$
then refer to limit comparison test.
answered Dec 1 at 21:39
gimusi
90.3k74495
thanks ill give it a try
– M.Jordan
Dec 1 at 21:42
@M.Jordan What about $sum frac1{n^frac32}$? Does it converge or diverge?
– gimusi
Dec 1 at 21:43
that would be a p series so it would coverge since its greater than 1 ?
– M.Jordan
Dec 1 at 21:44
Yes exactly $sum frac1{n^p}$ converges for any $p>1$ and the given series behaves as that for $n$ large. To formalize that in a proof by LCT we need to show that $$lim_{nto infty} frac{frac{sqrt n (n+1)}{(n^2+1)(n-1)}}{frac1{n^frac32}}=lim_{nto infty}frac{n^frac32sqrt n (n+1)}{(n^2+1)(n-1)}=Lin mathbb{R}$$
– gimusi
Dec 1 at 21:47
add a comment |
up vote
1
down vote
up vote
1
down vote
HINT
We have that
$$frac{sqrt n (n+1)}{(n^2+1)(n-1)} sim frac{n^frac32}{n^3}=frac1{n^frac32}$$
then refer to limit comparison test.
answered Dec 1 at 21:39
gimusi
90.3k74495
HINT
We have that
$$frac{sqrt n (n+1)}{(n^2+1)(n-1)} sim frac{n^frac32}{n^3}=frac1{n^frac32}$$
then refer to limit comparison test.
answered Dec 1 at 21:39
gimusi
90.3k74495
edited Dec 1 at 21:42
answered Dec 1 at 21:39
gimusi
90.3k74495
answered Dec 1 at 21:39
gimusi
90.3k74495
answered Dec 1 at 21:39
gimusi
90.3k74495
90.3k74495
thanks ill give it a try
– M.Jordan
Dec 1 at 21:42
@M.Jordan What about $sum frac1{n^frac32}$? Does it converge or diverge?
– gimusi
Dec 1 at 21:43
that would be a p series so it would coverge since its greater than 1 ?
– M.Jordan
Dec 1 at 21:44
Yes exactly $sum frac1{n^p}$ converges for any $p>1$ and the given series behaves as that for $n$ large. To formalize that in a proof by LCT we need to show that $$lim_{nto infty} frac{frac{sqrt n (n+1)}{(n^2+1)(n-1)}}{frac1{n^frac32}}=lim_{nto infty}frac{n^frac32sqrt n (n+1)}{(n^2+1)(n-1)}=Lin mathbb{R}$$
– gimusi
Dec 1 at 21:47
add a comment |
thanks ill give it a try
– M.Jordan
Dec 1 at 21:42
@M.Jordan What about $sum frac1{n^frac32}$? Does it converge or diverge?
– gimusi
Dec 1 at 21:43
that would be a p series so it would coverge since its greater than 1 ?
– M.Jordan
Dec 1 at 21:44
Yes exactly $sum frac1{n^p}$ converges for any $p>1$ and the given series behaves as that for $n$ large. To formalize that in a proof by LCT we need to show that $$lim_{nto infty} frac{frac{sqrt n (n+1)}{(n^2+1)(n-1)}}{frac1{n^frac32}}=lim_{nto infty}frac{n^frac32sqrt n (n+1)}{(n^2+1)(n-1)}=Lin mathbb{R}$$
– gimusi
Dec 1 at 21:47
thanks ill give it a try
– M.Jordan
Dec 1 at 21:42
thanks ill give it a try
– M.Jordan
Dec 1 at 21:42
@M.Jordan What about $sum frac1{n^frac32}$? Does it converge or diverge?
– gimusi
Dec 1 at 21:43
@M.Jordan What about $sum frac1{n^frac32}$? Does it converge or diverge?
– gimusi
Dec 1 at 21:43
that would be a p series so it would coverge since its greater than 1 ?
– M.Jordan
Dec 1 at 21:44
that would be a p series so it would coverge since its greater than 1 ?
– M.Jordan
Dec 1 at 21:44
Yes exactly $sum frac1{n^p}$ converges for any $p>1$ and the given series behaves as that for $n$ large. To formalize that in a proof by LCT we need to show that $$lim_{nto infty} frac{frac{sqrt n (n+1)}{(n^2+1)(n-1)}}{frac1{n^frac32}}=lim_{nto infty}frac{n^frac32sqrt n (n+1)}{(n^2+1)(n-1)}=Lin mathbb{R}$$
– gimusi
Dec 1 at 21:47
Yes exactly $sum frac1{n^p}$ converges for any $p>1$ and the given series behaves as that for $n$ large. To formalize that in a proof by LCT we need to show that $$lim_{nto infty} frac{frac{sqrt n (n+1)}{(n^2+1)(n-1)}}{frac1{n^frac32}}=lim_{nto infty}frac{n^frac32sqrt n (n+1)}{(n^2+1)(n-1)}=Lin mathbb{R}$$
– gimusi
Dec 1 at 21:47
add a comment |
up vote
1
down vote
We have
$$
frac{n+1}{n-1}=frac{n-1+2}{n-1}=1+frac{2}{n-1}le 3
$$
Also
$$
frac{1}{n^2+1}<frac{1}{n^2}
$$
Hence, for $nge2$,
$$
0<frac{sqrt{n}(n+1)}{(n^2+1)(n-1)}<frac{3}{n^{3/2}}
$$
Can you finish?
answered Dec 1 at 22:05
egreg
175k1383198
add a comment |
up vote
1
down vote
We have
$$
frac{n+1}{n-1}=frac{n-1+2}{n-1}=1+frac{2}{n-1}le 3
$$
Also
$$
frac{1}{n^2+1}<frac{1}{n^2}
$$
Hence, for $nge2$,
$$
0<frac{sqrt{n}(n+1)}{(n^2+1)(n-1)}<frac{3}{n^{3/2}}
$$
Can you finish?
answered Dec 1 at 22:05
egreg
175k1383198
add a comment |
up vote
1
down vote
up vote
1
down vote
We have
$$
frac{n+1}{n-1}=frac{n-1+2}{n-1}=1+frac{2}{n-1}le 3
$$
Also
$$
frac{1}{n^2+1}<frac{1}{n^2}
$$
Hence, for $nge2$,
$$
0<frac{sqrt{n}(n+1)}{(n^2+1)(n-1)}<frac{3}{n^{3/2}}
$$
Can you finish?
answered Dec 1 at 22:05
egreg
175k1383198
We have
$$
frac{n+1}{n-1}=frac{n-1+2}{n-1}=1+frac{2}{n-1}le 3
$$
Also
$$
frac{1}{n^2+1}<frac{1}{n^2}
$$
Hence, for $nge2$,
$$
0<frac{sqrt{n}(n+1)}{(n^2+1)(n-1)}<frac{3}{n^{3/2}}
$$
Can you finish?
answered Dec 1 at 22:05
egreg
175k1383198
answered Dec 1 at 22:05
egreg
175k1383198
answered Dec 1 at 22:05
egreg
175k1383198
answered Dec 1 at 22:05
egreg
175k1383198
175k1383198
add a comment |
add a comment |
Use D'Alembert criterion and compute $lim_{ntoinfty}frac{a_{n+1}}{a_n}$
– Tito Eliatron
Dec 1 at 21:39
2
Welcome to MSE! If possible, you should learn to use MathJax so that you can type your problem into the question rather than giving a link: math.meta.stackexchange.com/questions/5020/…
– Frpzzd
Dec 1 at 21:39