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Determine whether the series is convergent or divergent.
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where is the series?
\[\sum_{n=1}^{\infty} (\frac{ 5 }{ n^4 }+\frac{ 4 }{ n \sqrt{n} } )\]
convergent \[1/n^{p} \] if p > 1 which in this case it is\[1/n^{4} and 1/n^{3/2}\]
I am sorry but I am not understanding
meh soz was in a bit of a rush ;) \[\sum_{0}^{\infty} \frac{ 1}{ n^{p} } \] is the P-series which is convergent for p>1 and divergent for p<1 or p=1 The series you wrote is just 1/n^4 + 1/n^(3/2) and since both 1/n^4 and 1/n^(3/2) are convergent so is their sum wanted to write it more technically but the equation menu kept bugging out so this will have to suffice :(
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Thank you!!
np
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