# Infinite series without k-1

1. Apr 28, 2010

### oddjobmj

1. The problem statement, all variables and given/known data
I'm told to evaluate the following to the thousandths place:

$$\infty$$
$$\Sigma$$ 7*(0.35)^k
k=1

2. Relevant equations
We know that an infinite equation can be expressed as:

S$$\infty$$=(a1)/1-rn

3. The attempt at a solution

The first term (a1) is 7 and r=.35 so I can plug those into the above equation. I can see that we're doing something along the lines of:

7+7(.35)+7(.35^2)+7(.35^3)

However, there are two issues I'm having.

1) How do I plug in n when it's infinity?

2) In the non-infinite sum of geometric series problems I've worked where in the $$\Sigma$$ equation it's r^k versus the normal format of r^k-1 I had to modify the problem to make it k-1 and ended up multiplying the whole thing by r and then subtracting r from the whole thing... It was kind of crazy and I'm still not sure why and what happened.

Do I have to do something like that here?

How do I solve this equation?

2. Apr 28, 2010

### HallsofIvy

No, this is wrong. The formula for a finite geometric sum is
$$\sum_{k=0}^n ar^k= \frac{a(1- r^{n+ 1})}{1- r}$$.

You get the formula for the infinite series by taking the limit as n goes to infinity. It only converges if -1< r< 1 and in that case $r^n$ goes to 0. The sum of an infinite geometric series,
$$\sum_{n=0}^\infty ar^n= \frac{a}{1- r}$$

There is no "n" in the formula for the sum of an infinite series.

No, you do not.

Use the formula
$$\sum_{n=0}^\infty ar^n= \frac{a}{1- r}$$

3. Apr 28, 2010

### oddjobmj

Alright, thanks!

Now I guess my issue is that k=1 in my equation and I'm really not sure how to get it to 0.

4. Apr 28, 2010

### oddjobmj

Just for future reference if anyone has this question again.

I figured it out after HallsofIvy's help, thanks!

You just take the result of a/(1-r) which happens to be 10.769 in this case and subtract 1*a or '7' to get the end result of 3.769.

5. Apr 29, 2010

### HallsofIvy

The "k= 0" term in $7(0.35)^k$ is $7(0.35)^0= 7$. To get the sum from k= 1 to infinity. ust use the formula from 0 to infinity, the subtract 7.

Or: let n= k-1 so that when k= 1, n= 0. Then k= n+ 1 so [itex]7(0.35)^k= 7(0.35)^{n+1}= 7(0.35)(0.35)^n[/math]. Use the formula from n=0 to infinity with [math]a_0= 7(0.35)= 2.45 and r still 0.35.

Both of those should give the same answer.