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I am asked to find the sum to the following infinite geometric series:

$\sum_{n=1}^{\infty}\frac{(2)(3^{n+1})}{5^n}$

I then factor out the 2 and one 3 from the $3^{n+1}$ and get:

$\sum_{n=1}^{\infty}6 (\frac{3}{5})^n$

this results in a = 6 and r = $\frac{3}{5}$ and therefore I found the sum to be:

$\frac {6}{1 - \frac{3}{5}} = 15$

I thought this to be correct, however in the answer key for some reason it redefines the limit of the sum such that:

$\sum_{n=1}^{\infty}6 (\frac{3}{5})^n = -6 + \sum_{n=0}^{\infty}6 (\frac{3}{5})^n $

which returns the sum of 9 instead. Why does he does he redefine the bounds of the summation like this? Are these answers essentially the same?

Siméon
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  • I understand that but why does he start it at 1? –  Dec 15 '14 at 21:06
  • The series you are given at the beginning starts at $1$; you weren't "lucky" enough to be given a series starting at $0$, so you have to do a little work to get it in that form. – BaronVT Dec 15 '14 at 21:14
  • For instance, what if I tell you that "the sum of the first $10$ positive integers $(1,2,\dots,10)$ is $55$", and then ask you what the sum of $3$ through $10$ is? You can use the information I told you, but you have to modify it: in this case, you know the sum is $55$ minus the numbers you don't have, i.e. $55 - 1 - 2 = 52$. – BaronVT Dec 15 '14 at 21:16

1 Answers1

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The geometric series formula you have is for a series starting at $n = 0$. If you want to apply that formula, you have to add and subtract that initial term, which is $6$.

That is, you have, whether you like it or not,

$$\sum_{n=1}^{\infty}6 (\frac{3}{5})^n$$

Your formula only works if the sum starts at $n = 0$. So, we fix that:

$$\sum_{n=1}^{\infty}6 (\frac{3}{5})^n = -6 + 6 + \sum_{n=1}^{\infty}6 (\frac{3}{5})^n = -6 + \sum_{n=0}^{\infty}6 (\frac{3}{5})^n $$

BaronVT
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