The given p-series 1 + (1/[tex]\sqrt[4]{2^{3} }[/tex]) + (1/[tex]\sqrt[4]{3^{3} }[/tex]) + (1/ [tex]\sqrt[4]{4^{3} }[/tex])+ (1/ [tex]\sqrt[4]{5^{3} }[/tex]) is divergent as the value of is equal to 3/4 ⇒p < 1.
As given in the question,
Given p-series is equal to :
1 + (1/[tex]\sqrt[4]{2^{3} }[/tex]) + (1/[tex]\sqrt[4]{3^{3} }[/tex]) + (1/ [tex]\sqrt[4]{4^{3} }[/tex])+ (1/ [tex]\sqrt[4]{5^{3} }[/tex])
As value of 1³ = 1 and fourth root of 1³ is equal to 1.
We can substitute 1 = (1 /fourth root of 1³) which is equal to
= (1/ [tex]\sqrt[4]{1^{3} }[/tex] ) + (1/[tex]\sqrt[4]{2^{3} }[/tex]) + (1/[tex]\sqrt[4]{3^{3} }[/tex]) + (1/ [tex]\sqrt[4]{4^{3} }[/tex])+ (1/ [tex]\sqrt[4]{5^{3} }[/tex])
Apply nth formula we get,
= [tex]\sum\limits^\infty_0 {\frac{1}{\sqrt[4]{n^{3} } } }[/tex]
⇒ p = 3/4
And 3/4 < 1
⇒ p < 1
⇒P-series is divergent.
Therefore, as the value of p =3/4<1 the given series 1 + (1/[tex]\sqrt[4]{2^{3} }[/tex]) + (1/[tex]\sqrt[4]{3^{3} }[/tex]) + (1/ [tex]\sqrt[4]{4^{3} }[/tex])+ (1/ [tex]\sqrt[4]{5^{3} }[/tex]) is divergent.
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