Finding the sum of an infinite series

[haflanagan]

I have the series of (-1)^(n-1) * (7/(6^(n-1))) from n=1 to infinity.
I have to find whether or not the series converges, and if it does converge, I have to find the sum.
I have no clue what I am doing here. My best guess was to look at the absolute value of the series. If that converges, then the original converges absolutely. If it diverges, alternating series test, blah blah blah.
Well, I found that the series converges (geometric, R<1). Now, I need to find the sum. How do I do this?? If I type the absolute value of the series (which is just 7/(6^n-1)) into my calculator, I get that it is equivalent to 42*6^-n which is equivalent to 42(1/6)^n which is a geometric series. And the sum of a converging geometric series is A/(1-R). In this case, A = 42, R = 1/6. So the sum should equal 252/5. Well...it doesn't. The correct answer is 6.
What am I doing wrong? My final is tomorrow so this is pretty urgent .
And sorry if I did something wrong in terms of posting this thread, I have never used this website before.
Thanks in advance!

 

[pka]

I have the series of (-1)^(n-1) * (7/(6^(n-1))) from n=1 to infinity.
I have to find whether or not the series converges, and if it does converge, I have to find the sum.

The series is \(\displaystyle 7\sum\limits_{n = 1}^\infty {\left( {\dfrac{{ - 1}}{6}} \right)^{n - 1} } \). The answer is 6.

 

[haflanagan]

Okay...I see that, thanks
Now what if the series is something like 9^n * 13^(n-1)?
Sorry, I'm not trying to get other people to do my homework for me or anything, it probably seems like that.

 

[pka]

Okay...I see that, thanks
Now what if the series is something like 9^n * 13^(n-1)?

\(\displaystyle 13\sum\limits_{n = 1}^\infty {\left( {\dfrac{{ 9}}{13}} \right)^{n} } \)

 

[lookagain]

I have the series of (-1)^(n-1) * (7/(6^(n-1))) from n=1 to infinity.
I have to find whether or not the series converges, and if it does converge, I have to find the sum.
I have no clue what I am doing here.

My best guess was to look at the absolute value of the series.
You are not looking at "the absolute value of the series." You are looking at the sum
of the series where each summand (terms that are added) is the absolute value of the
respective term in the original series.

If that converges, then the original converges absolutely. If it diverges, alternating series test,
blah blah blah.
Well, I found that the series converges (geometric, R<1). Now, I need to find the sum.
How do I do this?? If I type the absolute value of the series (which is just 7/(6^(n-1))
into my calculator, I get that it is equivalent to 42*6^(-n) which is equivalent
to 42(1/6)^n which is a geometric series. And the sum of a converging geometric series
is A/(1-R). In this case, A = 42, R = 1/6. So the sum should equal 252/5. Well...it doesn't.
The correct answer is 6.
Why would you expect that the sum of the absolute values of all of the terms of the
original series would equal the sum of the original series, given that at least one term
in the original series is negative?

...

 

[lookagain]

Now what if the series is something like > > 9^n * 13^(n-1) < < ?

That is not a series.


\(\displaystyle \displaystyle\sum_{n = 1}^\infty \ (9^n)(13^{n - 1}) \ \ is \ \ a \ \ series \ \ and \ \ is \ \ divergent.\)



\(\displaystyle \displaystyle\sum_{n = 1}^\infty \ \dfrac{9^n}{13^{n - 1}} \ \ is \ \ also \ \ a \ \ series \ \ and \ \ is \ \ convergent.\)


\(\displaystyle \displaystyle\sum_{n = 1}^\infty \ \dfrac{9^n}{13^{n - 1}} \ = \)


\(\displaystyle 9\displaystyle\sum_{n = 1}^\infty \ \dfrac{9^{n - 1}}{13^{n - 1}} \ = \)


\(\displaystyle 9\displaystyle\sum_{n = 1}^\infty \ \bigg(\dfrac{9}{13}\bigg)^{n - 1} \ \)


For a/(1 - r) here, a = 1 and 1 - r = 1 - 9/13 = 4/13


a/(1 - r) = 1/(4/13) = 13/4


So, \(\displaystyle \displaystyle\sum_{n = 1}^\infty \ \dfrac{9^n}{13^{n - 1}} \ = \)

\(\displaystyle 9\bigg(\dfrac{13}{4}\bigg) \ = \ \dfrac{117}{4}.\)