I know the following function is convergent, but I'm not sure what the comparison function is.
FInd the comparison function of Summation from n=1 to infinity of (5+(-1)^n)/(n*n^(1/3))
I appreciate your help
FInd the comparison function of Summation from n=1 to infinity of (5+(-1)^n)/(n*n^(1/3))
I appreciate your help
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Note that (5 + (-1)^n) / (n*n^(1/3)) always has positive terms since 5 + (-1)^n ≥ 5 - 1 = 4.
So, we can use the Comparison Test.
Next, (5 + (-1)^n) / (n*n^(1/3))
= (5 + (-1)^n) / n^(4/3)
≤ (5 + 1) / n^(4/3)
= 6/n^(4/3); this is your comparison function.
Since Σ(n = 1 to ∞) 6/n^(4/3) is convergent (being a multiple of a convergent p-series with p = 4/3 > 1), we conclude that the series in question also converges.
I hope this helps!
So, we can use the Comparison Test.
Next, (5 + (-1)^n) / (n*n^(1/3))
= (5 + (-1)^n) / n^(4/3)
≤ (5 + 1) / n^(4/3)
= 6/n^(4/3); this is your comparison function.
Since Σ(n = 1 to ∞) 6/n^(4/3) is convergent (being a multiple of a convergent p-series with p = 4/3 > 1), we conclude that the series in question also converges.
I hope this helps!