Find the sum of the series if it converges: Summation as n=1 to inf 14/(-6)^n
\[\sum_{n=1}^{\infty} \space \space \space \frac{ 14 }{ (-6^{n}) }\]
We need to try and look at the series, is that maybe a special series type?
Lets look at some of the terms for it. Can you try and plug in 1 for the n in the expression for the An and set up the A1, the first term? Without doing the calculations, just try and write it out, so we can see the first one and then a few more after that
(-7/3)+7/18+(-7/108)+7/648+(-7/3888)+...+
my idea is to work a limiting ratio (a_[n+1])/(an)
that tells us if it converges, and then we can use the geometric sum formula as long as r < 1
The lag is real ;-;
since |r| = 1/6 the terms limit to 0 so it converges
Lag is the least of my worries - the open question scroll bar bled into this side of the window, so it's very difficult to read these responses... My professor did a very poor job of going over these. First lecture on these was a video he emailed us, and the second one was scatterbrained. He's too focused on his March Madness to teach :/
lol I know how you feel, some professors these days nothing much you can do about them
\(\color{blue}{\text{Originally Posted by}}\) @amistre64 since |r| = 1/6 the terms limit to 0 so it converges \(\color{blue}{\text{End of Quote}}\) and now we can apply the formula for it -- (formula for the sum of a geometric series)
S=a/1-r would become S=(-7/3)/1-(1/6), right?
-14(1/6)^1 +14(1/6)^2 -14(1/6)^3 +14(1/6)^4 -14(1/6)^5 +14(1/6)^6 if r = (1/6)^2 we may be able to detemrine this with 2 sets
its an alternating series, so we have to be catious is all
14(1)/(1--1/6) might work but im leary of it
7(14)/6 maybe?
the formula is for values of n from 0 and in your case you have it from n = 1
let the first term be -14/6 then
(-14/6)/(1-(-1/6)) = -2 Like that?
id feel better with splitting it into 2 parts, negative and positive
|dw:1427766344842:dw| what we can do is to add and subtract the term for n = 0
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