# Thread: Determine whether the following series converges or diverges.

1. ## Determine whether the following series converges or diverges.

Determine whether the following series converges or diverges

∑ [(-3/∏)^n + (e/9)^(n+1)]
n=1

If it converges, find its sum.

I don't know how to go about solving this one, can someone explain the steps that need to be taken?

2. ## Re: Determine whether the following series converges or diverges.

Distribute the sum over both terms, and if the series converges, both of those terms should converge.

If ∑an and ∑bn are convergent series, then so are the series ∑can (where c is a constant)

Then ∑(an + bn) =
an + ∑bn

3. ## Re: Determine whether the following series converges or diverges.

Originally Posted by sjmiller
Distribute the sum over both terms, and if the series converges, both of those terms should converge.

If ∑an and ∑bn are convergent series, then so are the series ∑can (where c is a constant)

Then ∑(an + bn) =
an + ∑bn
Im still not sure where to go from there. This is a geometric series, correct?

4. ## Re: Determine whether the following series converges or diverges.

Originally Posted by sjmiller
Distribute the sum over both terms, and if the series converges, both of those terms should converge.

If ∑an and ∑bn are convergent series, then so are the series ∑can (where c is a constant)

Then ∑(an + bn) =
an + ∑bn
Originally Posted by Preston019
Im still not sure where to go from there. This is a geometric series, correct?
Correct they are both geometric series. Remember the sum of such a series ∑ arn-1 = a/(1-r) provided that |r|<1
You need to get them into a form where you can easily determine the values of a and r. If |r| > 1 then the series diverges.

5. ## Re: Determine whether the following series converges or diverges.

Originally Posted by sjmiller
Correct they are both geometric series. Remember the sum of such a series ∑ arn-1 = a/(1-r) provided that |r|<1
You need to get them into a form where you can easily determine the values of a and r. If |r| > 1 then the series diverges.
Okay, so is this true: ∑ ar^(-n-1) = a/(1-r)^-1

And how would I get ∑(-3/∏)^n into the proper form?

6. ## Re: Determine whether the following series converges or diverges.

Originally Posted by Preston019
Okay, so is this true: ∑ ar^(-n-1) = a/(1-r)^-1

And how would I get ∑(-3/∏)^n into the proper form?
No that does not follow from the original formula.

Note that (-3/pi)^n can be written as (-3/pi)(-3/pi)^(n-1) where a = (-3/pi) and r = (-3/pi) and |r| < 1 so it indeed converges.

7. ## Re: Determine whether the following series converges or diverges.

Originally Posted by sjmiller
No that does not follow from the original formula.

Note that (-3/pi)^n can be written as (-3/pi)(-3/pi)^(n-1) where a = (-3/pi) and r = (-3/pi) and |r| < 1 so it indeed converges.
So, then how do I put ∑(e/9)^(n+1) into the proper form?

8. ## Re: Determine whether the following series converges or diverges.

Originally Posted by Preston019
Determine whether the following series converges or diverges

∑ [(-3/∏)^n + (e/9)^(n+1)]
n=1
If it converges, find its sum.
Lets do the using partial sums.
If $S_n = \sum\limits_{k = 1}^n {\left[ {\left( {\frac{{ - 3}}{\pi }} \right)^k + \left( {\frac{e}{9}} \right)^k } \right]}$ would you agree that $S_n=U_n+V_n$ where $U_n = \sum\limits_{k = 1}^n {\left[ {\left( {\frac{{ - 3}}{\pi }} \right)^k } \right]}$ and $V_n = \sum\limits_{k = 1}^n { {\left( {\frac{e}{9}} \right)^k } }$

Does each of $U_n~\&~V_n$ converge? If so, the sum of their limits is the sum of the original series.

9. ## Re: Determine whether the following series converges or diverges.

Originally Posted by Preston019
So, then how do I put ∑(e/9)^(n+1) into the proper form?
Same idea as the prior, instead of decreasing n by 1 we now want to decrease n by 2.

(e/9)^(n-1) = (e/9)^2(e/9)^(n-1)

Not that if you add the exponents together you get 2 + (n - 1) = n + 1
This is always your goal with a geometric series. Always check that the absolute value of r is less than 1, because if r is greater than 1 the series diverges.