Absolute or conditional convergence?
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0
down vote
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Determine whether the series:
$$sum_{n=1}^infty (-1)^n frac {2n^2+3n+4} {2n^4 + 3}$$
converges absolutely, conditionally or diverges.
I know the series converges conditionally using alternating series test.
My question is how do I determine absolute convergence here.
I tried limit comparison test with $frac{1}{n^2}$ which results in conditional convergence only, is that the way to go or not?
calculus sequences-and-series absolute-convergence conditional-convergence
asked Nov 26 at 23:11
J. Lastin
735
add a comment |
up vote
0
down vote
favorite
Determine whether the series:
$$sum_{n=1}^infty (-1)^n frac {2n^2+3n+4} {2n^4 + 3}$$
converges absolutely, conditionally or diverges.
I know the series converges conditionally using alternating series test.
My question is how do I determine absolute convergence here.
I tried limit comparison test with $frac{1}{n^2}$ which results in conditional convergence only, is that the way to go or not?
calculus sequences-and-series absolute-convergence conditional-convergence
asked Nov 26 at 23:11
J. Lastin
735
1
How did your comparison test result in only conditional convergence?
– Arthur
Nov 26 at 23:14
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Determine whether the series:
$$sum_{n=1}^infty (-1)^n frac {2n^2+3n+4} {2n^4 + 3}$$
converges absolutely, conditionally or diverges.
I know the series converges conditionally using alternating series test.
My question is how do I determine absolute convergence here.
I tried limit comparison test with $frac{1}{n^2}$ which results in conditional convergence only, is that the way to go or not?
calculus sequences-and-series absolute-convergence conditional-convergence
asked Nov 26 at 23:11
J. Lastin
735
Determine whether the series:
$$sum_{n=1}^infty (-1)^n frac {2n^2+3n+4} {2n^4 + 3}$$
converges absolutely, conditionally or diverges.
I know the series converges conditionally using alternating series test.
My question is how do I determine absolute convergence here.
I tried limit comparison test with $frac{1}{n^2}$ which results in conditional convergence only, is that the way to go or not?
calculus sequences-and-series absolute-convergence conditional-convergence
calculus sequences-and-series absolute-convergence conditional-convergence
asked Nov 26 at 23:11
J. Lastin
735
asked Nov 26 at 23:11
J. Lastin
735
asked Nov 26 at 23:11
J. Lastin
735
asked Nov 26 at 23:11
J. Lastin
735
asked Nov 26 at 23:11
J. Lastin
735
735
1
How did your comparison test result in only conditional convergence?
– Arthur
Nov 26 at 23:14
add a comment |
1
How did your comparison test result in only conditional convergence?
– Arthur
Nov 26 at 23:14
1
1
How did your comparison test result in only conditional convergence?
– Arthur
Nov 26 at 23:14
How did your comparison test result in only conditional convergence?
– Arthur
Nov 26 at 23:14
add a comment |
1 Answer
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up vote
3
down vote
accepted
Since$$leftlvert(-1)^nfrac{2n^2+3n+4}{2n^4+3}rightrvert=frac{2n^2+3n+4}{2n^4+3}$$and since$$lim_{ntoinfty}frac{frac{2n^2+3n+4}{2n^4+3}}{frac1{n^2}}=1,$$one deduces from the comparaison test that the series converges absolutely.
answered Nov 26 at 23:17
José Carlos Santos
147k22117217
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Since$$leftlvert(-1)^nfrac{2n^2+3n+4}{2n^4+3}rightrvert=frac{2n^2+3n+4}{2n^4+3}$$and since$$lim_{ntoinfty}frac{frac{2n^2+3n+4}{2n^4+3}}{frac1{n^2}}=1,$$one deduces from the comparaison test that the series converges absolutely.
answered Nov 26 at 23:17
José Carlos Santos
147k22117217
add a comment |
up vote
3
down vote
accepted
Since$$leftlvert(-1)^nfrac{2n^2+3n+4}{2n^4+3}rightrvert=frac{2n^2+3n+4}{2n^4+3}$$and since$$lim_{ntoinfty}frac{frac{2n^2+3n+4}{2n^4+3}}{frac1{n^2}}=1,$$one deduces from the comparaison test that the series converges absolutely.
answered Nov 26 at 23:17
José Carlos Santos
147k22117217
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Since$$leftlvert(-1)^nfrac{2n^2+3n+4}{2n^4+3}rightrvert=frac{2n^2+3n+4}{2n^4+3}$$and since$$lim_{ntoinfty}frac{frac{2n^2+3n+4}{2n^4+3}}{frac1{n^2}}=1,$$one deduces from the comparaison test that the series converges absolutely.
answered Nov 26 at 23:17
José Carlos Santos
147k22117217
Since$$leftlvert(-1)^nfrac{2n^2+3n+4}{2n^4+3}rightrvert=frac{2n^2+3n+4}{2n^4+3}$$and since$$lim_{ntoinfty}frac{frac{2n^2+3n+4}{2n^4+3}}{frac1{n^2}}=1,$$one deduces from the comparaison test that the series converges absolutely.
answered Nov 26 at 23:17
José Carlos Santos
147k22117217
answered Nov 26 at 23:17
José Carlos Santos
147k22117217
answered Nov 26 at 23:17
José Carlos Santos
147k22117217
answered Nov 26 at 23:17
José Carlos Santos
147k22117217
147k22117217
add a comment |
add a comment |
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1
How did your comparison test result in only conditional convergence?
– Arthur
Nov 26 at 23:14