0
$\begingroup$

If $$f(x) = \sum_{n=1}^\infty \frac{2^n}{n}x^n \: \: \text{for}\: \: x \in ]-\frac{1}{2}, \frac{1}{2}[$$ then what is $f'(x)$?

Attempt

It turns out that $\rho = \frac{1}{2}$ is the radius of convergence for this series. Since $f(x)$ has the form $\sum_{n=1}^\infty c_nx^n$, where $c_n = \frac{2^n}{n}$, we know that $$f'(x) = \sum_{n=1}^\infty c_nnx^{n-1} $$ Plugging in $c_n$ I get

$$f'(x) = \sum_{n=1}^\infty 2^nx^{n-1} = \sum_{n=1}^\infty 2 \cdot 2^{n-1} \cdot x^{n-1} = 2\sum_{n=1}^\infty (2x)^{n-1} $$

I feel I'm close, but I can't figure out what the next step should be.

$\endgroup$
1
  • 1
    $\begingroup$ $\sum_{k=0}^\infty q^k=\frac{1}{1-q}$ for suitable $q$. $\endgroup$
    – P. Quinton
    Commented Aug 11, 2022 at 12:17

1 Answer 1

Reset to default
2
$\begingroup$

Hint: Geometric series.

$$f'(x)=2\sum_{n=0}^\infty (2x)^n = \frac{2}{1-2x}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .