# Need help finding the limit of geometric serires

I'm learning some series tests in calculus and I can't completely figure this out. I know it's easier than i'm making it. Here's the question:

Determine whether the geometric series is divergent or convergent. If it is convergent, evaluate its limit. If it diverges state your answer as DIV.

$2/3-(2/3)^3+(2/3)^5-(2/3)^7$

I know that it's convergent. But i'm unsure of how to find the limit. Do I have to find the common ratio between them?

My teacher is saying I have to find the "$n$-th term" first. Which is just $(2/3)^n$

• If the $n$-th term is $(2/3)^n$, then your series would be $(2/3)^1+(2/3)^2+(2/3)^3+(2/3)^4+\cdots$. Your actual series is different, however. – vadim123 Apr 23 '14 at 20:12
• Hint: $2/3 - ({2/3})^3 + ({2/3})^5 - ({2/3})^7 + ... = 2/3(1 +(16/81) + (16/81)^2 +...) - (8/27)(1 +(16/81) + (16/81)^2 +...)$. – user3294068 Apr 23 '14 at 20:22

You have a difference between the sum to infinity of two geometric series, both with ratio $\left(\frac{2}{3}\right)^4$:- $$\sum_{k=0}^\infty\left(\frac{2}{3}\right)^{4k+1}-\sum_{k=0}^\infty\left(\frac{2}{3}\right)^{4k+3}$$ For a geometric series with first term $a$ and ratio $r$, the sum to infinity is given by formula $$S_\infty=\frac{a}{1-r}$$ Thus we have $$\sum_{k=0}^\infty\left(\frac{2}{3}\right)^{4k+1}-\sum_{k=0}^\infty\left(\frac{2}{3}\right)^{4k+3}=\frac{\frac{2}{3}}{1-\left(\frac{2}{3}\right)^4}-\frac{\left(\frac{2}{3}\right)^3}{1-\left(\frac{2}{3}\right)^4}=\frac{6}{13}$$

• =6/13 :) ${}{}{}{}{}$ – Matt L. Apr 23 '14 at 20:24
• Yes indeed, and many thanks. Will edit answer. – Alijah Ahmed Apr 23 '14 at 20:25

Your $n$-th term is

$$\left(\frac{2}{3}\right)^{2n+1}(-1)^n=\frac{2}{3}\left(-\frac{4}{9}\right)^{n}$$

$$\frac{2}{3}\sum_{n=0}^{\infty}\left(-\frac{4}{9}\right)^{n}$$

If you know the formula

$$\sum_{n=0}^{\infty}q^n=\frac{1}{1-q}$$

then you're done.

The two numbers you need for an geometric series are the first term, $a$, and the common ratio, $r$.

If $|r|<1$ then the sum to infinity is given by the formula $$S_{\infty} = \frac{a}{1-r}$$

Clearly $a=\frac{2}{3}$ in your example, but what is the common ratio $r$? What number do you need to multiply each term to get the next term?

Once you have $r$, use the formula for $S_{\infty}$.

I suspect you've been asked to find the $n^{\text{th}}$-term to use the ratio test to prove convergence. A geometric series looks like $a+ar+ar^2+ar^3+\cdots$ and so the $n^{\text{th}}$-term is $U_n = ar^{n-1}$.

In your case, applying the formula for the summation of geometric series, we see that the sum of the first $n$ terms is $q \frac{-q^{2n}-1}{-q^2-1}$ (where $q= \frac{2}{3}$) That's because the first term is $q$ and the denumenator is $-q^2$. Now observe that $q^2n$ approaches zero, therefore the limit is $\frac{q}{q^2+1}$.