Looking for assistance (some step by steps, hopefully, not just the answer) with the following homework problem:
Determine whether the series is absolutely convergent:
∑ ((n^2 + 1) / (2n^2 + 1))^2n, where n = 1, to infinity
It's giving me fits so thanks so much in advance for any assistance!
Determine whether the series is absolutely convergent:
∑ ((n^2 + 1) / (2n^2 + 1))^2n, where n = 1, to infinity
It's giving me fits so thanks so much in advance for any assistance!
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oo
∑ ((n^2 + 1) / (2n^2 + 1))^2n
n=1
It seems like the whole fraction is raised to the 2n.
Use the Root Test.
L = lim n-->oo | [(n^2 + 1) / (2n^2 + 1)]^(2n) | ^(1/n)
L = lim n-->oo | (n^2 + 1)^2 / (2n^2 + 1)^2 |
L = lim n-->oo | (n^4 + 2n^2 + 1) / (4n^4 + 4n^2 + 1) |
L = lim n-->oo | (1 + 2/n^2 + 1/n^4) / (4 + 4/n^2 + 1/n^4) |
L = 1/4.
Since 1/4 < 1, the series converges absolutely and hence converges.
Hope this helped.
∑ ((n^2 + 1) / (2n^2 + 1))^2n
n=1
It seems like the whole fraction is raised to the 2n.
Use the Root Test.
L = lim n-->oo | [(n^2 + 1) / (2n^2 + 1)]^(2n) | ^(1/n)
L = lim n-->oo | (n^2 + 1)^2 / (2n^2 + 1)^2 |
L = lim n-->oo | (n^4 + 2n^2 + 1) / (4n^4 + 4n^2 + 1) |
L = lim n-->oo | (1 + 2/n^2 + 1/n^4) / (4 + 4/n^2 + 1/n^4) |
L = 1/4.
Since 1/4 < 1, the series converges absolutely and hence converges.
Hope this helped.
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No problem. Math is a passion of mine, anytime.
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