# Math Help - [SOLVED] Finding The Sum Of This Series

1. ## [SOLVED] Finding The Sum Of This Series

I've attached the problem I'm stuck on, I tried using the formula for geometric series several times but I'm guessing that because there are two terms I should be going about it a different way? Could someone help me out?

2. Originally Posted by fattydq
I've attached the problem I'm stuck on, I tried using the formula for geometric series several times but I'm guessing that because there are two terms I should be going about it a different way? Could someone help me out?
Based on the answer you gave, it looks like you were close. The key thing to notice is that the exponent on $\frac{1}{2}$ is $n-1$. Thus, you sum is equivalent to $\sum_{n=0}^{\infty}7\left(\frac{1}{2}\right)^n$

The rest of the solution:
Spoiler:

$= 7\cdot\frac{1}{1-\frac{1}{2}} = 14$

3. Originally Posted by redsoxfan325
Based on the answer you gave, it looks like you were close. The key thing to notice is that the exponent on $\frac{1}{2}$ is $n-1$. Thus, you sum is equivalent to $\sum_{n=0}^{\infty}7\left(\frac{1}{2}\right)^n$

The rest of the solution:
Spoiler:

$= 7\cdot\frac{1}{1-\frac{1}{2}} = 14$
How can you just change n to 0? You can do that?

4. I also changed the exponent from $n-1$ to $n$. In the previous sum, $n$ starts at $1$ so $n-1$ starts at $0$. In this sum, both $n$ and the exponent start at $0$. It's the same. If you have doubts, try writing out the first few terms, and you'll see that they're the same.

5. Hello, fattydq!

Find the sum of the series if it converges: . $\sum^{\infty}_{n=1}7\left(\tfrac{1}{2}\right)^{n-1}$

... but I'm guessing that because there are two terms ... . .
What two terms?

We have: . $S \;=\;7 + 7\left(\tfrac{1}{2}\right) + 7\left(\tfrac{1}{2}\right)^2 + 7\left(\tfrac{1}{2}\right)^3 + \hdots$

This is an infinite geometric series with first term $a = 7$ and common ratio $r = \tfrac{1}{2}$

Its sum is: . $S \;=\;\frac{7}{1 - \frac{1}{2}} \;=\;14$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

That second problem is interesting . . .

$\sum^{\infty}_{n=0}\frac{(6\pi)^n}{18^{n+1}}$

We have: . $\frac{(6\pi)^n}{18^{n+1}} \;=\;\frac{(6\pi)^n}{18\cdot18^n} \;=\; \frac{1}{18}\cdot\frac{(6\pi)^n}{18^n} \;=\;\frac{1}{18}\left(\frac{6\pi}{18}\right)^n \;=\;\frac{1}{18}\left(\frac{\pi}{3}\right)^n$

And the series is: . $S \;=\;\frac{1}{18} + \frac{1}{18}\left(\frac{\pi}{3}\right) + \frac{1}{18}\left(\frac{\pi}{3}\right)^2 + \frac{1}{18}\left(\frac{\pi}{3}\right)^3 +\hdots$

This too is an infinite geometric series,

. . but since $r \,=\,\tfrac{\pi}{3}\,>\,1$, the series diverges.