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Tanya [424]
2 years ago
14

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 ''
Mathematics
1 answer:
inna [77]2 years ago
6 0

Best guess for the function is

\displaystyle f(x) = \sum_{n=1}^\infty \frac{x^n}{n^2}

By the ratio test, the series converges for

\displaystyle \lim_{n\to\infty} \left|\frac{x^{n+1}}{(n+1)^2} \cdot \frac{n^2}{x^n}\right| = |x| \lim_{n\to\infty} \frac{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 <em>closed</em> interval \boxed{-1 \le x \le 1}.

The derivative of f is the series

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

which also converges for |x| by the ratio test:

\displaystyle \lim_{n\to\infty} \left|\frac{x^{n+1}}{n+1} \cdot \frac n{x^n}\right| = |x| \lim_{n\to\infty} \frac{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 <em>closed-open</em> interval \boxed{-1 \le x < 1}.

Differentiating f once more gives the series

\displaystyle f''(x) = \sum_{n=1}^\infty \frac{n(n-1)x^{n-2}}{n^2} = \frac1{x^2} \sum_{n=1}^\infty \frac{(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|, endpoints not included.

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

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

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