Absolute or conditional convergence?
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
1025
add a comment |
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
1025
1
How did your comparison test result in only conditional convergence?
– Arthur
Nov 26 at 23:14
add a comment |
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
1025
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
1025
asked Nov 26 at 23:11
J. Lastin
1025
asked Nov 26 at 23:11
J. Lastin
1025
asked Nov 26 at 23:11
J. Lastin
1025
asked Nov 26 at 23:11
J. Lastin
1025
1025
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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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
149k22117219
add a comment |
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1 Answer
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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
149k22117219
add a comment |
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
149k22117219
add a comment |
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
149k22117219
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
149k22117219
answered Nov 26 at 23:17
José Carlos Santos
149k22117219
answered Nov 26 at 23:17
José Carlos Santos
149k22117219
answered Nov 26 at 23:17
José Carlos Santos
149k22117219
149k22117219
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