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How to evaluate $lim_{n to infty} sum_{k=1}^n frac{n+k}{n^3+k}$?
up vote
0
down vote
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How to evaluate the following limit?
$$lim_{n to infty} sum_{k=1}^n frac{n+k}{n^3+k}$$
I think that every term of the sum is greater than the first one and smaller than the last one and then from the squeeze theorem the limit is $0$. However, I can't prove that inequality.
calculus sequences-and-series limits
edited 2 days ago
user587192
1,21810
asked Nov 19 at 18:27
user69503
486
add a comment |
up vote
0
down vote
favorite
How to evaluate the following limit?
$$lim_{n to infty} sum_{k=1}^n frac{n+k}{n^3+k}$$
I think that every term of the sum is greater than the first one and smaller than the last one and then from the squeeze theorem the limit is $0$. However, I can't prove that inequality.
calculus sequences-and-series limits
edited 2 days ago
user587192
1,21810
asked Nov 19 at 18:27
user69503
486
1
What are you asking?
– Will M.
Nov 19 at 18:31
1
The sum is less than $n frac{n +n}{n^3+1}$
– RRL
Nov 19 at 18:33
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How to evaluate the following limit?
$$lim_{n to infty} sum_{k=1}^n frac{n+k}{n^3+k}$$
I think that every term of the sum is greater than the first one and smaller than the last one and then from the squeeze theorem the limit is $0$. However, I can't prove that inequality.
calculus sequences-and-series limits
edited 2 days ago
user587192
1,21810
asked Nov 19 at 18:27
user69503
486
How to evaluate the following limit?
$$lim_{n to infty} sum_{k=1}^n frac{n+k}{n^3+k}$$
I think that every term of the sum is greater than the first one and smaller than the last one and then from the squeeze theorem the limit is $0$. However, I can't prove that inequality.
calculus sequences-and-series limits
calculus sequences-and-series limits
edited 2 days ago
user587192
1,21810
asked Nov 19 at 18:27
user69503
486
edited 2 days ago
user587192
1,21810
asked Nov 19 at 18:27
user69503
486
edited 2 days ago
user587192
1,21810
edited 2 days ago
user587192
1,21810
edited 2 days ago
user587192
1,21810
1,21810
asked Nov 19 at 18:27
user69503
486
asked Nov 19 at 18:27
user69503
486
asked Nov 19 at 18:27
user69503
486
486
1
What are you asking?
– Will M.
Nov 19 at 18:31
1
The sum is less than $n frac{n +n}{n^3+1}$
– RRL
Nov 19 at 18:33
add a comment |
1
What are you asking?
– Will M.
Nov 19 at 18:31
1
The sum is less than $n frac{n +n}{n^3+1}$
– RRL
Nov 19 at 18:33
1
1
What are you asking?
– Will M.
Nov 19 at 18:31
What are you asking?
– Will M.
Nov 19 at 18:31
1
1
The sum is less than $n frac{n +n}{n^3+1}$
– RRL
Nov 19 at 18:33
The sum is less than $n frac{n +n}{n^3+1}$
– RRL
Nov 19 at 18:33
add a comment |
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
This should work:
$$sum_{k=1}^nfrac{n+k}{n^3+k}leqsum_{k=1}^nfrac{n+k}{n^3}leqsum_{k=1}^nfrac{n+n}{n^3}=frac{2n^2}{n^3}=frac{2}{n}to0$$
Also, every term is $geq 0$ so $0$ is also a lower bound.
answered Nov 19 at 18:33
Marco
1909
add a comment |
up vote
1
down vote
HINT:
Note that $n+1le n+kle 2n$ and $n^3+1le n^3 +kle n^3+n$.
answered Nov 19 at 18:35
Mark Viola
129k1273170
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
This should work:
$$sum_{k=1}^nfrac{n+k}{n^3+k}leqsum_{k=1}^nfrac{n+k}{n^3}leqsum_{k=1}^nfrac{n+n}{n^3}=frac{2n^2}{n^3}=frac{2}{n}to0$$
Also, every term is $geq 0$ so $0$ is also a lower bound.
answered Nov 19 at 18:33
Marco
1909
add a comment |
up vote
5
down vote
accepted
This should work:
$$sum_{k=1}^nfrac{n+k}{n^3+k}leqsum_{k=1}^nfrac{n+k}{n^3}leqsum_{k=1}^nfrac{n+n}{n^3}=frac{2n^2}{n^3}=frac{2}{n}to0$$
Also, every term is $geq 0$ so $0$ is also a lower bound.
answered Nov 19 at 18:33
Marco
1909
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
This should work:
$$sum_{k=1}^nfrac{n+k}{n^3+k}leqsum_{k=1}^nfrac{n+k}{n^3}leqsum_{k=1}^nfrac{n+n}{n^3}=frac{2n^2}{n^3}=frac{2}{n}to0$$
Also, every term is $geq 0$ so $0$ is also a lower bound.
answered Nov 19 at 18:33
Marco
1909
This should work:
$$sum_{k=1}^nfrac{n+k}{n^3+k}leqsum_{k=1}^nfrac{n+k}{n^3}leqsum_{k=1}^nfrac{n+n}{n^3}=frac{2n^2}{n^3}=frac{2}{n}to0$$
Also, every term is $geq 0$ so $0$ is also a lower bound.
answered Nov 19 at 18:33
Marco
1909
answered Nov 19 at 18:33
Marco
1909
answered Nov 19 at 18:33
Marco
1909
answered Nov 19 at 18:33
Marco
1909
1909
add a comment |
add a comment |
up vote
1
down vote
HINT:
Note that $n+1le n+kle 2n$ and $n^3+1le n^3 +kle n^3+n$.
answered Nov 19 at 18:35
Mark Viola
129k1273170
add a comment |
up vote
1
down vote
HINT:
Note that $n+1le n+kle 2n$ and $n^3+1le n^3 +kle n^3+n$.
answered Nov 19 at 18:35
Mark Viola
129k1273170
add a comment |
up vote
1
down vote
up vote
1
down vote
HINT:
Note that $n+1le n+kle 2n$ and $n^3+1le n^3 +kle n^3+n$.
answered Nov 19 at 18:35
Mark Viola
129k1273170
HINT:
Note that $n+1le n+kle 2n$ and $n^3+1le n^3 +kle n^3+n$.
answered Nov 19 at 18:35
Mark Viola
129k1273170
answered Nov 19 at 18:35
Mark Viola
129k1273170
answered Nov 19 at 18:35
Mark Viola
129k1273170
answered Nov 19 at 18:35
Mark Viola
129k1273170
129k1273170
add a comment |
add a comment |
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1
What are you asking?
– Will M.
Nov 19 at 18:31
1
The sum is less than $n frac{n +n}{n^3+1}$
– RRL
Nov 19 at 18:33