Question

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 the convergence intervall?

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 the convergence intervall?

Answer 1

The series:

`P(x) = sum_(n=1)^oo x^(n+2)/n`

converges for `x in [-1,1)` and its sum is:

`P(x) = x^2 ln abs (1-x)`

Explanation:

Consider the series:

`P(x) = sum_(n=1)^oo x^(n+2)/n`

Applying the ratio test we have:

`lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) abs ( ( x^(n+3)/(n+1)) / (x^(n+2)/n))`

`lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) abs ( (n/(n+1)) ( x^(n+3)/ (x^(n+2)))`

`lim_(n->oo) abs (a_(n+1)/a_n) = abs x lim_(n->oo) n/(n+1) = absx`

The series is then absolutely convergent for `abs x < 1` that is for `x in (-1,1)` and not convergent for `absx >1`

For `x=1` the series is the harmonic series that we know to be divergent, and for `x=-1` is the alternate harmonic series that can be proven to be convergent using Leibniz' theorem.

In conclusion the series is convergent in the interval `x in [-1,1)`.

Note now that:

`P(x) = sum_(n=1)^oo x^(n+2)/n = x^2sum_(n=1)^oo x^n/n`

Using the product rule:

`P'(x) = 2x sum_(n=1)^oo x^n/n +x^2 d/dx ( sum_(n=1)^oo x^n/n)`

Inside this interval then the series can be differentiated term by term, obtaining a series with the same radius of convergence, so:

`P'(x) = 2x sum_(n=1)^oo x^n/n +x^2 sum_(n=1)^oo d/dx ( x^n/n)`

`P'(x) = 2x sum_(n=1)^oo x^n/n +x^2 sum_(n=1)^oo x^(n-1)`

Divide an multiply the first sum by `x^2` and change the index of the second sum to `k=n-1`:

`P'(x) = 2/x sum_(n=1)^oo x^(n+2)/n +x^2 sum_(k=0)^oo x^k`

So the first sum is `P(x)` itself and the second sum is a geometric series and for `x in (-1,1)` we have:

`sum_(k=0)^oo x^k = 1/(1-x)`

`P'(x) = 2/x P(x) + x^2/(1-x)`

or:

`P'(x) - 2/x P(x) = x^2/(1-x)`

We can also note that starting from the geometric series, and always in the interior of the interval of convergence:

`sum_(k=0)^oo x^k = 1/(1-x)`

integrating term by term:

`sum_(k=0)^oo int_0^x t^kdt = int_0^x dt/(1-t)`

`sum_(k=0)^oo x^(k+1)/(k+1) = ln abs(1-x)`

change the index to `n = k+1`

`sum_(n=1)^oo x^n/n = ln abs(1-x)`

and multiplying by `x^2`:

`sum_(n=1)^oo x^(n+2)/n = x^2 ln abs(1-x)`