find the point of convergence of sequence {$a_n$} [duplicate]
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This question already has an answer here:
Convergence of a sequence, $a_n=sum_1^nn/(n^2+k)$
1 answer
Question about $a_n=sum_{k=1}^n frac{n}{n^2+k} $for $ n inmathbb{N}$
2 answers
Let $displaystyle a_n= sum_{k=1}^{n} frac{n}{n^2+k}$, for $nin mathbb{N}$. Then what is the nature of sequence ${a_n}_{ninmathbb{N}}$.
I tried using the Cauchy's general principle of converges for a sequence. But I think that this won't help me as because:
$displaystyle a_{n+p}= sum_{k=1}^{n+p} frac{n+p}{{(n+p)}^2+k}$ and $displaystyle a_n= sum_{k=1}^{n} frac{n}{n^2+k}$
And now if I do $a_{n+p}-a_{n}$ then this won't even cancel a single term.
$a_1$ will have one term.
$a_2$ will have two terms, and so on.
But here the first term in $a_2$ is not the term of $a_1$.
and due to this problem I was unable to use any results of convergence of series of positive terms.
Any help/hint will be appreciated.
real-analysis sequences-and-series cauchy-sequences
edited Jan 9 at 6:49
El borito
664216
asked Jan 9 at 6:26
SinghSingh
1,18011134
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marked as duplicate by Martin R, Martin Sleziak, Cesareo, DRF, RRL real-analysis
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Jan 9 at 13:34
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 |
$begingroup$
This question already has an answer here:
Convergence of a sequence, $a_n=sum_1^nn/(n^2+k)$
1 answer
Question about $a_n=sum_{k=1}^n frac{n}{n^2+k} $for $ n inmathbb{N}$
2 answers
Let $displaystyle a_n= sum_{k=1}^{n} frac{n}{n^2+k}$, for $nin mathbb{N}$. Then what is the nature of sequence ${a_n}_{ninmathbb{N}}$.
I tried using the Cauchy's general principle of converges for a sequence. But I think that this won't help me as because:
$displaystyle a_{n+p}= sum_{k=1}^{n+p} frac{n+p}{{(n+p)}^2+k}$ and $displaystyle a_n= sum_{k=1}^{n} frac{n}{n^2+k}$
And now if I do $a_{n+p}-a_{n}$ then this won't even cancel a single term.
$a_1$ will have one term.
$a_2$ will have two terms, and so on.
But here the first term in $a_2$ is not the term of $a_1$.
and due to this problem I was unable to use any results of convergence of series of positive terms.
Any help/hint will be appreciated.
real-analysis sequences-and-series cauchy-sequences
edited Jan 9 at 6:49
El borito
664216
asked Jan 9 at 6:26
SinghSingh
1,18011134
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marked as duplicate by Martin R, Martin Sleziak, Cesareo, DRF, RRL real-analysis
Users with the real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.
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Jan 9 at 13:34
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.
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Also: math.stackexchange.com/q/77909, math.stackexchange.com/q/1995860 – all found with Approach0
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– Martin R
Jan 9 at 7:57
add a comment |
$begingroup$
This question already has an answer here:
Convergence of a sequence, $a_n=sum_1^nn/(n^2+k)$
1 answer
Question about $a_n=sum_{k=1}^n frac{n}{n^2+k} $for $ n inmathbb{N}$
2 answers
Let $displaystyle a_n= sum_{k=1}^{n} frac{n}{n^2+k}$, for $nin mathbb{N}$. Then what is the nature of sequence ${a_n}_{ninmathbb{N}}$.
I tried using the Cauchy's general principle of converges for a sequence. But I think that this won't help me as because:
$displaystyle a_{n+p}= sum_{k=1}^{n+p} frac{n+p}{{(n+p)}^2+k}$ and $displaystyle a_n= sum_{k=1}^{n} frac{n}{n^2+k}$
And now if I do $a_{n+p}-a_{n}$ then this won't even cancel a single term.
$a_1$ will have one term.
$a_2$ will have two terms, and so on.
But here the first term in $a_2$ is not the term of $a_1$.
and due to this problem I was unable to use any results of convergence of series of positive terms.
Any help/hint will be appreciated.
real-analysis sequences-and-series cauchy-sequences
edited Jan 9 at 6:49
El borito
664216
asked Jan 9 at 6:26
SinghSingh
1,18011134
$endgroup$
This question already has an answer here:
Convergence of a sequence, $a_n=sum_1^nn/(n^2+k)$
1 answer
Question about $a_n=sum_{k=1}^n frac{n}{n^2+k} $for $ n inmathbb{N}$
2 answers
Let $displaystyle a_n= sum_{k=1}^{n} frac{n}{n^2+k}$, for $nin mathbb{N}$. Then what is the nature of sequence ${a_n}_{ninmathbb{N}}$.
I tried using the Cauchy's general principle of converges for a sequence. But I think that this won't help me as because:
$displaystyle a_{n+p}= sum_{k=1}^{n+p} frac{n+p}{{(n+p)}^2+k}$ and $displaystyle a_n= sum_{k=1}^{n} frac{n}{n^2+k}$
And now if I do $a_{n+p}-a_{n}$ then this won't even cancel a single term.
$a_1$ will have one term.
$a_2$ will have two terms, and so on.
But here the first term in $a_2$ is not the term of $a_1$.
and due to this problem I was unable to use any results of convergence of series of positive terms.
Any help/hint will be appreciated.
This question already has an answer here:
Convergence of a sequence, $a_n=sum_1^nn/(n^2+k)$
1 answer
Question about $a_n=sum_{k=1}^n frac{n}{n^2+k} $for $ n inmathbb{N}$
2 answers
real-analysis sequences-and-series cauchy-sequences
real-analysis sequences-and-series cauchy-sequences
edited Jan 9 at 6:49
El borito
664216
asked Jan 9 at 6:26
SinghSingh
1,18011134
edited Jan 9 at 6:49
El borito
664216
asked Jan 9 at 6:26
SinghSingh
1,18011134
edited Jan 9 at 6:49
El borito
664216
edited Jan 9 at 6:49
El borito
664216
edited Jan 9 at 6:49
El borito
664216
664216
asked Jan 9 at 6:26
SinghSingh
1,18011134
asked Jan 9 at 6:26
SinghSingh
1,18011134
asked Jan 9 at 6:26
SinghSingh
1,18011134
1,18011134
marked as duplicate by Martin R, Martin Sleziak, Cesareo, DRF, RRL real-analysis
Users with the real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.
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Jan 9 at 13:34
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, Martin Sleziak, Cesareo, DRF, RRL real-analysis
Users with the real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.
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Jan 9 at 13:34
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.
$begingroup$
Also: math.stackexchange.com/q/77909, math.stackexchange.com/q/1995860 – all found with Approach0
$endgroup$
– Martin R
Jan 9 at 7:57
add a comment |
$begingroup$
Also: math.stackexchange.com/q/77909, math.stackexchange.com/q/1995860 – all found with Approach0
$endgroup$
– Martin R
Jan 9 at 7:57
$begingroup$
Also: math.stackexchange.com/q/77909, math.stackexchange.com/q/1995860 – all found with Approach0
$endgroup$
– Martin R
Jan 9 at 7:57
$begingroup$
Also: math.stackexchange.com/q/77909, math.stackexchange.com/q/1995860 – all found with Approach0
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– Martin R
Jan 9 at 7:57
add a comment |
2 Answers
2
active
oldest
votes
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Try sandwiching $a_n$.
$$a_n = sum_{k=1}^n frac{n}{n^2 + k} le sum_{k=1}^n frac{1}{n} = 1.$$
$$a_n = sum_{k=1}^n frac{n}{n^2 + k} ge sum_{k=1}^n frac{1}{n + 1} = frac{n}{n+1}.$$
answered Jan 9 at 6:31
angryavianangryavian
42.5k23481
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add a comment |
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Note that for $1le kle n$ :$${nover n^2+n}le {nover n^2+k}le {nover n^2+1}$$therefore $${n^2over n^2+n}le a_nle {n^2over n^2+1}$$and bu squeeze theorem$$lim_{nto infty}a_n=1$$
answered Jan 9 at 7:29
Mostafa AyazMostafa Ayaz
18.2k31040
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Already posted by another user, right?
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– Did
Jan 9 at 15:34
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Try sandwiching $a_n$.
$$a_n = sum_{k=1}^n frac{n}{n^2 + k} le sum_{k=1}^n frac{1}{n} = 1.$$
$$a_n = sum_{k=1}^n frac{n}{n^2 + k} ge sum_{k=1}^n frac{1}{n + 1} = frac{n}{n+1}.$$
answered Jan 9 at 6:31
angryavianangryavian
42.5k23481
$endgroup$
add a comment |
$begingroup$
Try sandwiching $a_n$.
$$a_n = sum_{k=1}^n frac{n}{n^2 + k} le sum_{k=1}^n frac{1}{n} = 1.$$
$$a_n = sum_{k=1}^n frac{n}{n^2 + k} ge sum_{k=1}^n frac{1}{n + 1} = frac{n}{n+1}.$$
answered Jan 9 at 6:31
angryavianangryavian
42.5k23481
$endgroup$
add a comment |
$begingroup$
Try sandwiching $a_n$.
$$a_n = sum_{k=1}^n frac{n}{n^2 + k} le sum_{k=1}^n frac{1}{n} = 1.$$
$$a_n = sum_{k=1}^n frac{n}{n^2 + k} ge sum_{k=1}^n frac{1}{n + 1} = frac{n}{n+1}.$$
answered Jan 9 at 6:31
angryavianangryavian
42.5k23481
$endgroup$
Try sandwiching $a_n$.
$$a_n = sum_{k=1}^n frac{n}{n^2 + k} le sum_{k=1}^n frac{1}{n} = 1.$$
$$a_n = sum_{k=1}^n frac{n}{n^2 + k} ge sum_{k=1}^n frac{1}{n + 1} = frac{n}{n+1}.$$
answered Jan 9 at 6:31
angryavianangryavian
42.5k23481
answered Jan 9 at 6:31
angryavianangryavian
42.5k23481
answered Jan 9 at 6:31
angryavianangryavian
42.5k23481
answered Jan 9 at 6:31
angryavianangryavian
42.5k23481
42.5k23481
add a comment |
add a comment |
$begingroup$
Note that for $1le kle n$ :$${nover n^2+n}le {nover n^2+k}le {nover n^2+1}$$therefore $${n^2over n^2+n}le a_nle {n^2over n^2+1}$$and bu squeeze theorem$$lim_{nto infty}a_n=1$$
answered Jan 9 at 7:29
Mostafa AyazMostafa Ayaz
18.2k31040
$endgroup$
$begingroup$
Already posted by another user, right?
$endgroup$
– Did
Jan 9 at 15:34
add a comment |
$begingroup$
Note that for $1le kle n$ :$${nover n^2+n}le {nover n^2+k}le {nover n^2+1}$$therefore $${n^2over n^2+n}le a_nle {n^2over n^2+1}$$and bu squeeze theorem$$lim_{nto infty}a_n=1$$
answered Jan 9 at 7:29
Mostafa AyazMostafa Ayaz
18.2k31040
$endgroup$
$begingroup$
Already posted by another user, right?
$endgroup$
– Did
Jan 9 at 15:34
add a comment |
$begingroup$
Note that for $1le kle n$ :$${nover n^2+n}le {nover n^2+k}le {nover n^2+1}$$therefore $${n^2over n^2+n}le a_nle {n^2over n^2+1}$$and bu squeeze theorem$$lim_{nto infty}a_n=1$$
answered Jan 9 at 7:29
Mostafa AyazMostafa Ayaz
18.2k31040
$endgroup$
Note that for $1le kle n$ :$${nover n^2+n}le {nover n^2+k}le {nover n^2+1}$$therefore $${n^2over n^2+n}le a_nle {n^2over n^2+1}$$and bu squeeze theorem$$lim_{nto infty}a_n=1$$
answered Jan 9 at 7:29
Mostafa AyazMostafa Ayaz
18.2k31040
edited Jan 9 at 7:49
answered Jan 9 at 7:29
Mostafa AyazMostafa Ayaz
18.2k31040
answered Jan 9 at 7:29
Mostafa AyazMostafa Ayaz
18.2k31040
answered Jan 9 at 7:29
Mostafa AyazMostafa Ayaz
18.2k31040
18.2k31040
$begingroup$
Already posted by another user, right?
$endgroup$
– Did
Jan 9 at 15:34
add a comment |
$begingroup$
Already posted by another user, right?
$endgroup$
– Did
Jan 9 at 15:34
$begingroup$
Already posted by another user, right?
$endgroup$
– Did
Jan 9 at 15:34
$begingroup$
Already posted by another user, right?
$endgroup$
– Did
Jan 9 at 15:34
add a comment |
$begingroup$
Also: math.stackexchange.com/q/77909, math.stackexchange.com/q/1995860 – all found with Approach0
$endgroup$
– Martin R
Jan 9 at 7:57