How do you test the alternating series #Sigma (-1)^nsqrtn/(n+1)# from n is #[1,oo)# for convergence?

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Answer 1 · 2017-02-02T16:30:49

The series:

#sum_(n=1)^oo (-1)^n sqrt(n)/(n+1)#

is convergent

The series:

#sum_(n=1)^oo (-1)^n sqrt(n)/(n+1)#

is an alternating series, so we can test its convergence using Leibniz's theorem, which states that an alternating series

#sum_(n=1)^oo (-1)^n a_n#

is convergent if:

(i) #lim_(n->oo) a_n = 0#

(ii) #a_(n+1) <= a_n#

in our case:

#lim_(n->oo) sqrt(n)/(n+1) = lim_(n->oo)1/((n+1)/ sqrt(n)) = lim_(n->oo)1/((sqrt(n) +1/ sqrt(n))) = 0#

so the first condition is satisfied.
For the second we analyze the function:

#f(x) = sqrt(x)/(x+1)#

and calculate the derivative:

#(df)/(dx) = ((x+1)/(2sqrt(x)) - sqrt(x))/(x+1)^2= (x+1-2x)/(2sqrt(x)(x+1)^2)= - (x-1)/(2sqrt(x)(x+1)^2)#

we can see that #(df)/(dx) < 0# for #x in (1,+oo)# therefore the function is strictly decreasing in that interval and we have:

#f(n+1) < f(n)#

that is:

#sqrt(n+1)/(n+2) <= sqrt(n)/(n+1)#

and also the second condition is satisfied.