Question

Calculus question about formulas with series

Answer 1

how does \[\frac{ 9*18*27...(9n) }{ n! }\]= \[9^n(n!)\]?

Answer 2

i would think that the formula = \[\frac{ 9n }{ n! }\]

Answer 3

\[\sum_{n=1}^{\infty}\] was supposed to be in front of that

Answer 4

`\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ f(n)}\)`
here is the code for sigma notation, if you want.

Answer 5

(it is an inline sigma )

Answer 6

Do you mean that it is like a set of terms

\(\large\color{black}{ \displaystyle \frac{9}{n!} }\) , \(\large\color{black}{ \displaystyle \frac{18}{n!} }\) , \(\large\color{black}{ \displaystyle \frac{27}{n!} }\) .....


then

\(\large\color{black}{ \displaystyle \frac{3 \times 3}{n!} }\) , \(\large\color{black}{ \displaystyle \frac{3 \times 6}{n!} }\) , \(\large\color{black}{ \displaystyle \frac{3 \times 9}{n!} }\) , \(\large\color{black}{ \displaystyle \frac{3 \times 12}{n!} }\) .....


and then,

\(\large\color{black}{ \displaystyle \frac{3 \times 3 \times 1}{n!} }\) , \(\large\color{black}{ \displaystyle \frac{3 \times 3 \times 2}{n!} }\) , \(\large\color{black}{ \displaystyle \frac{3 \times 3 \times 3}{n!} }\) , \(\large\color{black}{ \displaystyle \frac{3 \times 3 \times 4}{n!} }\) .....


then,

\(\large\color{black}{ \displaystyle \frac{9}{n!} }\) , \(\large\color{black}{ \displaystyle \frac{9\times 2}{n!} }\) , \(\large\color{black}{ \displaystyle \frac{9 \times 3}{n!} }\) , \(\large\color{black}{ \displaystyle \frac{9 \times 4}{n!} }\) .....


\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } \frac{9 \times n}{n!}}\)


YES, CORRECT !

Answer 7

hope this is making some sense:)

Answer 8

idk why but the formula you gave me above did not appear in letters for me. maybe my connection needs reset again

Answer 9

let me be more clear @SolomonZelman. The question states: Determine whether the series is absolutely convergent, conditionally convergent, or divergent and then gives me \[\sum_{n=1}^{\infty}\frac{ 9*18*27...(9n) }{ n! }\] When i ask for a hint, it tells me the limit= \[\lim_{n \rightarrow \infty}\frac{ 9^n(n!) }{ n! }\]

Answer 10

I dont understand how they get 9^n for that

Answer 11

Well.. the way I see it:
$$
\frac{9 \cdot 18 \cdot 27 .... (9n)}{n!} = \frac{(9_1 \cdot 1) \cdot (9_2 \cdot 2) \cdot (9_3 \cdot 3) .... (9_n \cdot n)}{n!} = \\
= \frac{(9_1 \cdot 9_2 \cdot 9_3 .... 9_n) \cdot (1\cdot2\cdot3 ... n)}{n!} = \frac{9^n \cdot n!}{n!} = 9^n
$$And that means that
$$
\sum_{n=1}^\infty ~ \frac{9 \cdot 18 \cdot 27 .... (9n)}{n!} = \sum_{n=1}^\infty ~ 9^n
$$which is a geometric series with |r|>1 so it's divergence.

Answer 12

Ok,... I can see that. but isnt, for example 9^2=81 and not 18?

Answer 13

Ye but the 18 comes from \((9 \cdot 9) \cdot (1 \cdot 2) =(1 \cdot 9) \cdot (2 \cdot 9) = 9 \cdot 18\)
It's still a chain of terms multiplied by each other, just reordered a little bit

Answer 14

oh wow

Answer 15

\( \Huge ☺\)

Answer 16

thank you

Answer 17

Sure, np