# converging/diverging of geometric series

• November 9th 2009, 04:54 PM
RhotelidlewildR
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.
• November 9th 2009, 04:58 PM
Drexel28
Quote:

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 $\sum_{n=1}^{\infty}\frac{(-6)^{n-1}}{5^{n-1}}$. Merely note that $\left|\frac{(-6)^{n-1}}{5^{n-1}}\right|\ge1$...so what?
• November 9th 2009, 05:00 PM
RhotelidlewildR
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.
• November 9th 2009, 05:41 PM
Drexel28
Quote:

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.
• November 9th 2009, 05:44 PM
RhotelidlewildR
thanks.
Quote:

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!
• November 9th 2009, 06:03 PM
Drexel28
Quote:

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 $\ln(n)$ come from? Also, if the series is indeed what I posted in my seminal post then the series diverges.
• November 9th 2009, 06:15 PM
RhotelidlewildR
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.
• November 9th 2009, 06:21 PM
JG89
If $\lim_{n \rightarrow \infty} \frac{(-6)^{n-1}}{5^{n-1}} \neq 0$ then $\sum_{i = 1}^{\infty} \frac{(-6)^{n-1}}{5^{n-1}}$ diverges.