How to find the sum of this Convergent series [duplicate]
This question already has an answer here:
Evaluate $sum_{n=1}^{infty} frac{n}{n^4+n^2+1}$
2 answers
I am given the following convergent series asked to find the value of:
$$sum_{n=1}^{infty} left(frac{4n}{(n^4+n^2 +1)}right)$$
According to Sum calculators, I have found the value to be 2. Though, I don't know what approach to take to prove this. To aid us in the question, we've been given the following hint:
$$n^4 + n^2 + 1 = (n^2 + 1)^2 − n^2$$
Can someone point me in the right direction?
sequences-and-series
asked Nov 26 '18 at 20:25
D4C
141
marked as duplicate by Martin R, Community♦ Nov 26 '18 at 20:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Evaluate $sum_{n=1}^{infty} frac{n}{n^4+n^2+1}$
2 answers
I am given the following convergent series asked to find the value of:
$$sum_{n=1}^{infty} left(frac{4n}{(n^4+n^2 +1)}right)$$
According to Sum calculators, I have found the value to be 2. Though, I don't know what approach to take to prove this. To aid us in the question, we've been given the following hint:
$$n^4 + n^2 + 1 = (n^2 + 1)^2 − n^2$$
Can someone point me in the right direction?
sequences-and-series
asked Nov 26 '18 at 20:25
D4C
141
marked as duplicate by Martin R, Community♦ Nov 26 '18 at 20:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Evaluate $sum_{n=1}^{infty} frac{n}{n^4+n^2+1}$
2 answers
I am given the following convergent series asked to find the value of:
$$sum_{n=1}^{infty} left(frac{4n}{(n^4+n^2 +1)}right)$$
According to Sum calculators, I have found the value to be 2. Though, I don't know what approach to take to prove this. To aid us in the question, we've been given the following hint:
$$n^4 + n^2 + 1 = (n^2 + 1)^2 − n^2$$
Can someone point me in the right direction?
sequences-and-series
asked Nov 26 '18 at 20:25
D4C
141
This question already has an answer here:
Evaluate $sum_{n=1}^{infty} frac{n}{n^4+n^2+1}$
2 answers
I am given the following convergent series asked to find the value of:
$$sum_{n=1}^{infty} left(frac{4n}{(n^4+n^2 +1)}right)$$
According to Sum calculators, I have found the value to be 2. Though, I don't know what approach to take to prove this. To aid us in the question, we've been given the following hint:
$$n^4 + n^2 + 1 = (n^2 + 1)^2 − n^2$$
Can someone point me in the right direction?
This question already has an answer here:
Evaluate $sum_{n=1}^{infty} frac{n}{n^4+n^2+1}$
2 answers
sequences-and-series
sequences-and-series
asked Nov 26 '18 at 20:25
D4C
141
asked Nov 26 '18 at 20:25
D4C
141
asked Nov 26 '18 at 20:25
D4C
141
asked Nov 26 '18 at 20:25
D4C
141
asked Nov 26 '18 at 20:25
D4C
141
141
marked as duplicate by Martin R, Community♦ Nov 26 '18 at 20:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Martin R, Community♦ Nov 26 '18 at 20:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
add a comment |
2 Answers
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$$n^4+n^2+1=$$
$$(n^2+n+1)(n^2-n+1)$$
$$frac{1}{n^2-n+1}-frac{1}{(n+1)^2-(n+1)+1}=$$
$$frac{2n}{n^4+n^2+1}$$
Think telescoping.
You will find $2$.
answered Nov 26 '18 at 20:34
hamam_Abdallah
37.9k21634
2
Before answering what is likely a duplicate question, search for similar questions already asked.
– amWhy
Nov 26 '18 at 20:42
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Further hint:
$$ frac{n}{n^4+n^2+1} = frac{n}{(n^2+1)^2-n^2} = frac{n}{(n^2+1-n)(n^2+1+n)} = frac{1}{2} left(frac{1}{n^2-n+1} - frac{1}{n^2+n+1}right). $$
answered Nov 26 '18 at 20:34
MisterRiemann
5,7941624
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$$n^4+n^2+1=$$
$$(n^2+n+1)(n^2-n+1)$$
$$frac{1}{n^2-n+1}-frac{1}{(n+1)^2-(n+1)+1}=$$
$$frac{2n}{n^4+n^2+1}$$
Think telescoping.
You will find $2$.
answered Nov 26 '18 at 20:34
hamam_Abdallah
37.9k21634
2
Before answering what is likely a duplicate question, search for similar questions already asked.
– amWhy
Nov 26 '18 at 20:42
add a comment |
$$n^4+n^2+1=$$
$$(n^2+n+1)(n^2-n+1)$$
$$frac{1}{n^2-n+1}-frac{1}{(n+1)^2-(n+1)+1}=$$
$$frac{2n}{n^4+n^2+1}$$
Think telescoping.
You will find $2$.
answered Nov 26 '18 at 20:34
hamam_Abdallah
37.9k21634
2
Before answering what is likely a duplicate question, search for similar questions already asked.
– amWhy
Nov 26 '18 at 20:42
add a comment |
$$n^4+n^2+1=$$
$$(n^2+n+1)(n^2-n+1)$$
$$frac{1}{n^2-n+1}-frac{1}{(n+1)^2-(n+1)+1}=$$
$$frac{2n}{n^4+n^2+1}$$
Think telescoping.
You will find $2$.
answered Nov 26 '18 at 20:34
hamam_Abdallah
37.9k21634
$$n^4+n^2+1=$$
$$(n^2+n+1)(n^2-n+1)$$
$$frac{1}{n^2-n+1}-frac{1}{(n+1)^2-(n+1)+1}=$$
$$frac{2n}{n^4+n^2+1}$$
Think telescoping.
You will find $2$.
answered Nov 26 '18 at 20:34
hamam_Abdallah
37.9k21634
answered Nov 26 '18 at 20:34
hamam_Abdallah
37.9k21634
answered Nov 26 '18 at 20:34
hamam_Abdallah
37.9k21634
answered Nov 26 '18 at 20:34
hamam_Abdallah
37.9k21634
37.9k21634
2
Before answering what is likely a duplicate question, search for similar questions already asked.
– amWhy
Nov 26 '18 at 20:42
add a comment |
2
Before answering what is likely a duplicate question, search for similar questions already asked.
– amWhy
Nov 26 '18 at 20:42
2
2
Before answering what is likely a duplicate question, search for similar questions already asked.
– amWhy
Nov 26 '18 at 20:42
Before answering what is likely a duplicate question, search for similar questions already asked.
– amWhy
Nov 26 '18 at 20:42
add a comment |
Further hint:
$$ frac{n}{n^4+n^2+1} = frac{n}{(n^2+1)^2-n^2} = frac{n}{(n^2+1-n)(n^2+1+n)} = frac{1}{2} left(frac{1}{n^2-n+1} - frac{1}{n^2+n+1}right). $$
answered Nov 26 '18 at 20:34
MisterRiemann
5,7941624
add a comment |
Further hint:
$$ frac{n}{n^4+n^2+1} = frac{n}{(n^2+1)^2-n^2} = frac{n}{(n^2+1-n)(n^2+1+n)} = frac{1}{2} left(frac{1}{n^2-n+1} - frac{1}{n^2+n+1}right). $$
answered Nov 26 '18 at 20:34
MisterRiemann
5,7941624
add a comment |
Further hint:
$$ frac{n}{n^4+n^2+1} = frac{n}{(n^2+1)^2-n^2} = frac{n}{(n^2+1-n)(n^2+1+n)} = frac{1}{2} left(frac{1}{n^2-n+1} - frac{1}{n^2+n+1}right). $$
answered Nov 26 '18 at 20:34
MisterRiemann
5,7941624
Further hint:
$$ frac{n}{n^4+n^2+1} = frac{n}{(n^2+1)^2-n^2} = frac{n}{(n^2+1-n)(n^2+1+n)} = frac{1}{2} left(frac{1}{n^2-n+1} - frac{1}{n^2+n+1}right). $$
answered Nov 26 '18 at 20:34
MisterRiemann
5,7941624
answered Nov 26 '18 at 20:34
MisterRiemann
5,7941624
answered Nov 26 '18 at 20:34
MisterRiemann
5,7941624
answered Nov 26 '18 at 20:34
MisterRiemann
5,7941624
5,7941624
add a comment |
add a comment |
