For which #x in RR# does the series P(x) converge and for which does it diverge? #P(x) = sum 1/n * x^(n+2)# And how do i show that the relation #P'(x) - 2*((P(x))/x) = x^2/(1-x)# is valid for the inner side of of the convergence intervall?

1 Answer
Mar 20, 2018

Convergence interval is #|x|<1# (I am assuming that the sum starts at #n=1#)

Explanation:

The #n#th term in the series above is given by

#t_n = x^{n+2}/n#

Thus we have

#t_{n+1}/t_n = x^{n+3}/{n+1}times n/x^{n+2} = n/{n+1}x#

so that

#lim_{n to oo} t_{n+1}/t_n = x#

and according to the ratio test, the series will converge for #|x|<1#.
(it is well known that for #x=1#, the series obtained, #1+1/2+1/3+...# diverges, while the series #-1+1/2-1/3+...#, obtained for #x=-1# is conditionally convergent).

Now, within the interval of convergence, a power series can be differentiated term by term, so that

#P'(x) = sum_{n=1}^oo 1/n times (n+2)x^{n+1} =sum_{n=1}^oo (1+2/n)x^{n+1} #

And thus

#P'(x) - 2(P(x))/x=sum_{n=1}^oo (1+2/n)x^{n+1} - 2 sum_{n=1}^oo 1/nx^{n+1} #
#qquad = sum_{n=1}^oo (1+2/n-2/n)x^{n+1}=sum_{n=1}^oo x^{n+1} #

The final result is an infinite geometric series with first term #x^2# and common ratio #x# - which obviously sums to #x^2/(1-x)# for #|x|<1#