Question

How would you solve the sum of a multiplicative series?

Answer 1

can u post the Q??

Answer 2

Getting to it, just a sec...

Answer 3

It says it's convergent, but I have no clue how to solve for that with this series.

Answer 4

GENRAL FORM OF THE SERIES
\[\sum_{1}^{\infty} (4n-1)/(9n-1)\]

Answer 5

But it looks like it's the sum of:

(the general form being multiplied from n=1 to infinity) + (the general form being multiplied from n=2 to infinity) + (the general form being multiplied from n=3 to infinity) + ... (the general form being multiplied from n=infinity to infinity)

Answer 6

if lim\[\lim_{n \rightarrow \infty} (n+1)/n <1\]

Answer 7

Okay... ratio test...

Answer 8

exactly bro

Answer 9

can u solve it now???

Answer 10

@LolGoCryAboutIt

Answer 11

Sorry, I was answering another question

I can try, hang on. (I usually have trouble with ratio test problems)

Answer 12

okkk buddy

Answer 13

\[\lim_{n \to \infty} \frac{a_{n+1}}{a_{n}} = L\]

Answer 14

\[\lim_{n \to \infty} \frac{\frac{4(n+1) -1}{9(n+1) - 1}}{\frac{4n -1}{9n - 1}} = L\]

Answer 15

\[\lim_{n \to \infty} \frac{(4(n+1)-1) \: \cdot (9n-1)}{(9(n+1)-1)\cdot(4n-1)}= L \]

Answer 16

\[\lim_{n \to \infty} \frac{(4n+3) \: \cdot (9n-1)}{(9n+8)\cdot(4n-1)}= L \]

Answer 17

\[\lim_{n \to \infty} \frac{((4n-1)+4) \: \cdot (9n-1)}{((9n-1)+9)\cdot(4n-1)}= L \]

Answer 18

\[\lim_{n \to \infty} \frac{4}{9} = L \]

Answer 19

\[\frac{4}{9} < 1\]
the series converges

Answer 20

Is that the correct way to do it @gorv?