Does $$\sum _{n=3}^{\infty }\:\frac{\pi \left(-3\right)^{n+1}}{7^n}$$ converge/diverge.
If it converges, state the sum.
I rewrote this series as follows:
$$\sum _{n=3}^{\infty }\:\frac{\pi \left(-3\right)^{n+1}}{7^n}$$
$$\sum _{n=3}^{\infty }\:\frac{\pi \left(-3\right)^{n+1}}{7^n}=\pi \sum _{n=3}^{\infty }\:\frac{\left(-3\right)^{n+1}}{7^n} = \pi \sum _{n=3}^{\infty }\:\frac{\left(-3\right)^{n}(-3)}{7^n} = (-3\pi) \sum _{n=3}^{\infty }\:(\frac{-3}{7})^n$$
Here we see the geometric series into play, in the form $\sum _{n=1}^{\infty }ar^n$, where $r$ = $\dfrac{-3}{7}$, and we know that if $|r| < 1$, then the series converges to $\dfrac{a}{1-r}$.
$a$ is first non-zero term, so we have $a = -3\pi (\dfrac{-3}{7})^3 = \dfrac{81\pi}{343}$
$\therefore$ the series converges to $\dfrac{\dfrac{81\pi}{343}}{1+\dfrac{3}{7}} \approx 0.52$
I think this is right, but I verified with Wolfram alpha, and it got different solutions. Who is correct?
I have no idea how you wrote it in WA, but this is the answer I'm getting: (I included my query for reference)