0
$\begingroup$

Using theorems about differentiation or integration of power series calculate infinite sum of

$$ \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)3^n} $$

The answer should equal to $\frac{\pi}{2\sqrt3}$.

I tried using $f(x) = \arctan(x) = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)}x^{2n+1}$ with $x=\frac{1}{3}$ but that fails, since we have $3^n$ and not $3^{2n+1}$ in the exponent.

$\endgroup$
2
$\begingroup$

Since$$\arctan(x)=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{2n+1},$$you have$$\frac\pi6=\arctan\left(\frac1{\sqrt3}\right)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)3^n\sqrt3}$$and therefore$$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)3^n}=\frac{\pi\sqrt3}6=\frac\pi{2\sqrt3}.$$

$\endgroup$
2
$\begingroup$

Factor an $x$ out entirely out of the sum, so you’re left with $x^{2n}$, then take $x=\frac1{\sqrt{3}}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.