Multiply these two series:

(1) $\displaystyle \Sigma^{\infty}_{n=1} \frac{1}{2^{n-1}}$

(2) $\displaystyle \Sigma^{\infty}_{n=1} \frac{1}{n\sqrt{n}}$

Printable View

- Jun 27th 2010, 11:37 AMAlso sprach ZarathustraMultiplication of two series...
Multiply these two series:

(1) $\displaystyle \Sigma^{\infty}_{n=1} \frac{1}{2^{n-1}}$

(2) $\displaystyle \Sigma^{\infty}_{n=1} \frac{1}{n\sqrt{n}}$ - Jun 27th 2010, 11:49 AMmfetch22
I believe the first series is equal to 2, correct? So wouldn't that imply that we just multiply each term in the second series by 2? I'm not certain, because I have no experience with multiplying infinite series, but wouldn't the result (since the first series is equal to 2) simply be:

$\displaystyle \Sigma^{\infty}_{n=1} \frac{2}{n\sqrt{n}}$

? - Jun 27th 2010, 12:30 PMAlso sprach Zarathustra
Oooops...