0
$\begingroup$

I just want to be sure I'm approaching this problem correctly. To find the sum of an infinite series, I usually first list out a few elements of the series by plugging in i. Then I find the common ratio between terms (r). Then I use the following formula to find the sum if it exists: $$\frac{a_1}{1-r}$$

I just feel like there is more to this one because the terms become so large when I sub in i for the exponents. I thought for a moment maybe this was a telescoping series but the terms don't cancel as I generate them. Is there a better way to approach this problem than the steps I mentioned above? I'm thinking maybe there is a trick like I should find a way to simplify this before I plug in i.

$$\sum_{i=4}^{\infty} \frac{5}{4^i} - \frac{3}{6^{i+1}}$$

$\endgroup$

2 Answers 2

0
$\begingroup$

Hint

$$\sum_{i=4}^{\infty}\left( \frac{5}{4^i} - \frac{3}{6^{i+1}}\right)=5\sum_{i=4}^{\infty}\frac{1}{4^i}-\frac 36\sum_{i=4}^{\infty}\frac{1}{6^i}=5\sum_{i=4}^{\infty}\frac{1}{4^i}-\frac 12\sum_{i=4}^{\infty}\frac{1}{6^i}$$

$\endgroup$
0
$\begingroup$

$$=\sum_{r=4}^n(f(r)-g(r))=\sum_{r=4}^nf(r)-\sum_{r=1}^ng(r)$$

When both the series sum converge to finite values

$\endgroup$

You must log in to answer this question.

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