# `sum_(n=1)^oo 1/(nroot(4)(n))` Determine the convergence or divergence of the series.

To evaluate the given series:` sum_(n=1)^oo 1/(nroot(4)n)` , we may apply  radical property: `root(a)(x^b) = x^(b/a)` and Law of exponent:` x^a*x^b= x^(a+b)` .

The given series becomes :

`sum_(n=1)^oo 1/(nroot(4)n) =sum_(n=1)^oo 1/(n*n^(1/4))`

` =sum_(n=1)^oo 1/n^(1/4+1)`

` =sum_(n=1)^oo 1/n^(5/4) `

or  `sum_(n=1)^oo 1/n^(1.25)`

The `sum_(n=1)^oo 1/n^(1.25) ` is in a form of a p-series.

In general, the p-series follows the following form:

`sum_(n=1)^oo 1/n^p =1/1^p + 1/2^p+ 1/3^p +1/4^p + 1/5^p +...`

Recall the theorem for a p-series states that `sum_(n=1)^oo 1/n^p` is convergent if `pgt1` and divergent if `plt=1` .

Applying the theorem on the given series `sum_(n=1)^oo 1/n^(1.25)` , we compare `n^(1.25)` with `n^p` to determine the corresponding value: `p =1.25` .

It satisfies `pgt1` since `1.25gt1` .

Therefore, the given series `sum_(n=1)^oo 1/(nroot(4)n)` is convergent.

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