## anonymous one year ago How do i find the sum?? http://i.imgur.com/tBwQ0Fi.png I know to represent it as a geometric series but I am unsure what to do with the ^n-1

1. Astrophysics

We can use geometric series test, so first lets find the partial sum for $\sum_{n=1}^{\infty} \frac{ (-4)^{n-1} }{ 5^n }$ so we let n = 1

2. Astrophysics

$a_1 = \frac{ (-4)^{1-1} }{ 5^1 } = \frac{ (-4)^0 }{ 5 } = \frac{ 1 }{ 5 }$ and we know if |r|<1 it is then convergent to find r we can use $a_1r=a_2 \implies \frac{ 1 }{ 5 }r=-\frac{ 4 }{ 25 } \implies r = -\frac{ 20 }{ 25 } = -\frac{ 4 }{ 5 }$ $\frac{ a_1 }{ 1-r } = \frac{ 1/5 }{ 1+4/5 } = \frac{ 1 }{ 9 }$ therefore it's convergent

3. Astrophysics

4. anonymous

Oh i see we manipulated a2/a1=r

5. Astrophysics

Yup

6. anonymous

Why did we use n=1?

7. Astrophysics

That gives us our first term a_1

8. Astrophysics

Remember r is the common ratio here

9. anonymous

Oh right n=1 in the sum so its our first term, i keep on thinking it starts at zero. Thanks

10. Astrophysics

Np :)