Calculus of a Single Variable, Chapter 9, 9.5, Section 9.5, Problem 41

To determine the convergence or divergence of the series sum_(n=1)^oo (-1)^n/sqrt(n) , we may apply the Root Test.
In Root test, we determine the limit as:
lim_(n-gtoo) root(n)(|a_n|)= L
or
lim_(n-gtoo) |a_n|^(1/n)= L
Then ,we follow the conditions:
a) L lt1 then the series converges absolutely
b) Lgt1 then the series diverges
c) L=1 or does not exist then the test is inconclusive.The series may be divergent, conditionally convergent, or absolutely convergent.
For the given series sum_(n=1)^oo (-1)^n/sqrt(n) , we have a_n =(-1)^n/sqrt(n).
Applying the Root test, we set-up the limit as:
lim_(n-gtoo) |(-1)^n/sqrt(n)|^(1/n) =lim_(n-gtoo) (1/sqrt(n))^(1/n) Note: |(-1)^n| = 1
Apply radical property: root(n)(x) =x^(1/n) and Law of exponent: (x/y)^n = x^n/y^n .
lim_(n-gtoo) (1/sqrt(n))^(1/n) =lim_(n-gtoo) (1/n^(1/2))^(1/n)
=lim_(n-gtoo) 1^(1/n) /n^(1/2*1/n)
=lim_(n-gtoo) 1^(1/n) /n^(1/(2n))
=lim_(n-gtoo) 1 /n^(1/(2n))
Apply the limit property: lim_(x-gta)[(f(x))/(g(x))] =(lim_(x-gta) f(x))/(lim_(x-gta) g(x)) .
lim_(n-gtoo) 1 /n^(1/(2n)) =(lim_(n-gtoo) 1 )/(lim_(n-gtoo)n^(1/(2n)))
= 1/1
=1
The limit value L = 1 implies that the series may be divergent, conditionally convergent, or absolutely convergent.
To verify, we use alternating series test on sum a_n .
a_n = 1/sqrt(n) is positive and decreasing from N=1
lim_(n-gtoo)1/sqrt(n) = 1/oo = 1
Based on alternating series test condition, the series sum_(n=1)^oo (-1)^n/sqrt(n) converges.
Apply p-series test on sum |a_n| .
sum_(n=1)^oo |(-1)^n/sqrt(n)|=sum_(n=1)^oo 1/sqrt(n).
=sum_(n=1)^oo 1/n^(1/2)
Based on p-series test condition, we have p=1/2 that satisfies 0ltplt=1 .
Thus, the series sum_(n=1)^oo |(-1)^n/sqrt(n)| diverges.
Conclusion:
sum_(n=1)^oo (-1)^n/sqrt(n) is conditionally convergent since sum_(n=1)^oo (-1)^n/sqrt(n) is convergent and sum_(n=1)^oo |(-1)^n/sqrt(n)| is divergent.