Evaluate each infinite series that has a sum. (I only need to do one) it's in the comments.

Question

bloomlocke367 · Answer

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

misty1212 · Answer

HI!!

misty1212 · Answer

use 
$$\frac{1}{1-r}$$and you are done in one step

misty1212 · Answer

you can do it in your head
what is one plus one third?

bloomlocke367 · Answer

4/3

misty1212 · Answer

yes 
and the reciprocal of 4/3?

bloomlocke367 · Answer

3/4

misty1212 · Answer

$$\color\magenta\heartsuit$$

bloomlocke367 · Answer

Why'd you do that, though?

bloomlocke367 · Answer

@misty1212

misty1212 · Answer

what does the question "why" mean in this context? does it mean why is 
$$\sum_{n=1}^{\infty}r^{n-1}=\frac{1}{1-r}$$?

misty1212 · Answer

you can derive the formula for yourself, and i am sure its derivation is in your book, it is not hard, but would be rather lengthy to write here

bloomlocke367 · Answer

yes. Where did $$\frac{ 1 }{ 1-r }$$ come from?

bloomlocke367 · Answer

wait, nevermind... it comes from$$S=\frac{ a_1 }{ 1-r}$$

misty1212 · Answer

that is how you sum a geometric series like this one
outline of proof like this, if you call the sum $S$
$$S=1+r+r^2+r^3+...\
rS=r+r^2+r^3+r^4+...\
S-rS=1\
S(1-r)=1\
S=\frac{1}{1-r}$$

preetha · Answer

Wow, very nicely done Misty!  I am impressed!

misty1212 · Answer

@Preetha thanks!

bloomlocke367 · Answer

My book says that if $$\left| r \right|<1$$ it has a finite sum and if$$\left| r \right|\ge1$$ it doesn't have a finite sum.. so, since r was -1/3, there's a sum?

bloomlocke367 · Answer

and that also means it converges, correct?

misty1212 · Answer

oh yes
that is because $$|-\frac{1}{3}|<1$$

bloomlocke367 · Answer

Okay. Thank you so much for your clarification!