How to determine convergence or divergence of series sigma(n=2, infinity ) #1/[sqrt(n)-1]# ?

Question

1791 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-13T02:43:14

Diverges by the Direct Comparison Test.

So, we have

#sum_(n=2)^oo1/(sqrtn-1)#

We can use the Direct Comparison Test here.

#a_n=1/(sqrtn-1)#, we must define some sequence #b_n# for comparison, and whose series convergence/divergence we can easily determine.

Well, if we omit the constantly subtracted #1# from the denominator,

#b_n=1/sqrtn<=a_n# for all #n,# as a larger denominator created by no subtraction of a constant entails a smaller overall sequence.

Now, we know

#sum_(n=2)^oo1/sqrtn=sum_(n=2)^oo1/n^(1/2)# diverges by the #p-#series test, as #p=1/2<1#.

Since the smaller series diverges, so must the larger series #sum_(n=2)^oo1/(sqrtn-1)#.

The P-series Test:

For a series in the form #sum_(n=a)^oo1/n^p,# we have convergence if #p>1# and divergence if #p<=1.#