Confused by Example 26: n=9, How Does k=0?

[Firepanda]

http://img214.imageshack.us/img214/3928/idontgetkm1.jpg

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?

 

[Firepanda]

Also,

http://img205.imageshack.us/img205/9089/cantix9.jpg

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?

 

[HallsofIvy]

http://img214.imageshack.us/img214/3928/idontgetkm1.jpg

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?

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
\sum_{j= 0}^9 \frac{2}{3}\frac{1}{3^j}[/itex]<br /> That is a geometric series with A= 2/3, r= 1/3, and n= 9.

 

[HallsofIvy]

Also,

http://img205.imageshack.us/img205/9089/cantix9.jpg

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?

For that one,
\lim_{x\rightarrow 0}\frac{e^x- x- 1}{sin^2(x)}[/itex]<br /> since both numerator and denominator go to 0 I would use L&#039;Hopital&#039;s rule (twice). And, again, you <b>don&#039;t</b> need to find the limit itself!