# Simple Series

1. Jan 17, 2008

### Firepanda

http://img214.imageshack.us/img214/3928/idontgetkm1.jpg [Broken]

For example 26, I don't see how (for the 1st part) n=9. In the definition of the summing of them it says n is the number above SIGMA, so why not 10? Is it to do with n=1, which isn't k=0 from the definition?

Last edited by a moderator: May 3, 2017
2. Jan 17, 2008

### Firepanda

Also,

http://img205.imageshack.us/img205/9089/cantix9.jpg [Broken]

I did the 1st, which diverged by the divergence test.

But the 2nd, I tried the ratio test, but it got far 2 complicated too early on. Which test do I use? I tried them all, what am I doing wrong?

Last edited by a moderator: May 3, 2017
3. Jan 17, 2008

### HallsofIvy

Staff Emeritus
Yes, the formula given for a geometric series starts with k= 0. The example given
$$\sum_{k= 1}^10 \frac{2}{3^n}$$
starts with k= 1. The simplest way to handle that is to factor 1/3 out of the product:
$$\sum_{k=1}^{10}\frac{2}{3} \frac{1}{3^{k-1}}$$
Now, let j= k-1. Then k= 1 becomes j= 1-1= 0 and k= 10 becomes j= $j= 10-1= 9$. Of course, $3^{k-1}$ becomes $3^j$. The sum is now
[tex]\sum_{j= 0}^9 \frac{2}{3}\frac{1}{3^j}[/itex]
That is a geometric series with A= 2/3, r= 1/3, and n= 9.

Last edited by a moderator: May 3, 2017
4. Jan 17, 2008

### HallsofIvy

Staff Emeritus
For that one,
[tex]\lim_{x\rightarrow 0}\frac{e^x- x- 1}{sin^2(x)}[/itex]
since both numerator and denominator go to 0 I would use L'Hopital's rule (twice). And, again, you don't need to find the limit itself!

Last edited by a moderator: May 3, 2017
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