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

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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.

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial