Question

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

Answer 1

The series:

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

is convergent

Explanation:

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.