# Thread: converging/diverging of geometric series

1. ## converging/diverging of geometric series

I have a question about the lim as it approaches infinity of ((-6)^(n-1))/(5^(n-1)). I am unsure what to do.. completely actually. I was thinking i would try to put it in a/(1-r) mode, but i dont really know how, or if that is even the right thing to do.

2. Originally Posted by RhotelidlewildR
I have a question about the lim as it approaches infinity of ((-6)^(n-1))/(5^(n-1)). I am unsure what to do.. completely actually. I was thinking i would try to put it in a/(1-r) mode, but i dont really know how, or if that is even the right thing to do.
You said series. So I assume you are talking about $\displaystyle \sum_{n=1}^{\infty}\frac{(-6)^{n-1}}{5^{n-1}}$. Merely note that $\displaystyle \left|\frac{(-6)^{n-1}}{5^{n-1}}\right|\ge1$...so what?

3. ## finding limits.

I need to find whether that is converging or diverging, and if convergence i need to find the limit. But yes, that is what i meant.

4. Originally Posted by RhotelidlewildR
I need to find whether that is converging or diverging, and if convergence i need to find the limit. But yes, that is what i meant.
But my dear RhotelidlewildR I did tell you whether or not it converges.

5. ## thanks.

Originally Posted by Drexel28
But my dear RhotelidlewildR I did tell you whether or not it converges.
yeah, so unless i am wrong (which i very well could be) it is greater than one so it converges. But my problem is breaking it down to what value it converges at. Do i divide by ln(n) since it is the loswest power in the numerator? I am just really lost. Thanks for your patience!

6. Originally Posted by RhotelidlewildR
yeah, so unless i am wrong (which i very well could be) it is greater than one so it converges. But my problem is breaking it down to what value it converges at. Do i divide by ln(n) since it is the loswest power in the numerator? I am just really lost. Thanks for your patience!
Now I am really lost. Where did this $\displaystyle \ln(n)$ come from? Also, if the series is indeed what I posted in my seminal post then the series diverges.

7. haha

i am sorry i accidentally ran off on another problem. So is this the way i do it... a=-1, r=(6/5), but if that is the case i end up with -1/-(1/5) or 5. this is confusing the life out of me.

8. If $\displaystyle \lim_{n \rightarrow \infty} \frac{(-6)^{n-1}}{5^{n-1}} \neq 0$ then $\displaystyle \sum_{i = 1}^{\infty} \frac{(-6)^{n-1}}{5^{n-1}}$ diverges.