Question

Let f(x) = [infinity] xn n2 n = 1. find the intervals of convergence for f. (enter your answers using interval notation. ) find the intervals of convergence for f '. find the intervals of convergence for f ''

Answer 1

Best guess for the function is

`\displaystyle f(x) = \sum_(n=1)^\infty (x^n)/(n^2)`

By the ratio test, the series converges for

`\displaystyle \lim_(n\to\infty) \left|(x^(n+1))/((n+1)^2) \cdot (n^2)/(x^n)\right| = |x| \lim_(n\to\infty) (n^2)/((n+1)^2) = |x| < 1`

When
`x=1`,
`f(x)` is a convergent
`p`-series.

When
`x=-1`,
`f(x)` is a convergent alternating series.

So, the interval of convergence for
`f(x)` is the closed interval
`\boxed{-1 \le x \le 1}`.

The derivative of
`f` is the series

`\displaystyle f'(x) = \sum_(n=1)^\infty (nx^(n-1))/(n^2) = \frac1x \sum_(n=1)^\infty \frac{x^n}n`

which also converges for
`|x|<1` by the ratio test:

`\displaystyle \lim_(n\to\infty) \left|(x^(n+1))/(n+1) \cdot \frac n{x^n}\right| = |x| \lim_(n\to\infty) (n)/(n+1) = |x| < 1`

When
`x=1`,
`f'(x)` becomes the divergent harmonic series.

When
`x=-1`,
`f'(x)` is a convergent alternating series.

The interval of convergence for
`f'(x)` is then the closed-open interval
`\boxed{-1 \le x < 1}`.

Differentiating
`f` once more gives the series

`\displaystyle f''(x) = \sum_(n=1)^\infty (n(n-1)x^(n-2))/(n^2) = \frac1{x^2} \sum_(n=1)^\infty ((n-1)x^n)/(n) = \frac1{x^2} \left(\sum_(n=1)^\infty x^n - \sum_(n=1)^\infty \frac{x^n}n\right)`

The first series is geometric and converges for
`|x|<1`, endpoints not included.

The second series is
`f'(x)`, which we know converges for
`-1\le x<1`.

Putting these intervals together, we see that
`f''(x)` converges only on the open interval
`\boxed{-1 < x < 1}`.