## blackstreet23 one year ago I am trying to do an alternating series test problem but i forgot how to prove if the series is decreasing. I remember they were several tests for that, but i forgot which where. Please help!

1. freckles

Can you show me your series?

2. freckles

Sometimes it is pretty easy with derivative. Sometimes it is easy with algebra. like we can show $a_{n+1}<a_n \text{ for }n \text{ using algebra } \\ \text{ or show } f' \text{ is negative for all } x$

3. freckles

n not x

4. blackstreet23

could you tell me all the tests? I am trying to review my calc 2 to be ready for calc 3

5. freckles

Those are the only two I can recall.

6. SolomonZelman

In alternating series, if the absolute value of the terms decreases then the series will converge.

7. blackstreet23

8. SolomonZelman

So, I will denote the sums with $$\rm S_1,~S_2...$$ |dw:1441594022749:dw|

9. blackstreet23

that is what i have so far

10. SolomonZelman

Can you post it in here, more clearly.....

11. freckles

Is the attachment bringing up a new question?

12. blackstreet23

no just showing my work so far

13. freckles

$k+1>k \text{ so } \frac{1}{k+1}<\frac{1}{k}$

14. freckles

or. ... $f(k)=\frac{1}{k+1} \\ f'(k)=\frac{-1}{(k+1)^2} \\ \frac{-1}{(k+1)^2}<0 \text{ for all } k \text{ except } k=-1 \\ \text{ which we aren't concerned about }$ either one of these ways shows a_k is a decreasing sequence

15. SolomonZelman

$$\large\color{black}{ \displaystyle \sum_{ k=0 }^{ \infty } \frac{x^k}{k+1}}$$ $$\large\color{black}{ \displaystyle \lim_{ k\rightarrow \infty }\frac{x^{k+1}(k+1)}{(k+2)x^k}}$$ $$\large\color{black}{ \displaystyle x\lim_{ k\rightarrow \infty }\frac{k+1}{k+2}}$$ $$\large\color{black}{ \displaystyle|x |<1 }$$

16. SolomonZelman

and it is actually good, because even if it was just x$$^k$$ without dividing by k+1, then it converges for |x|<1. And when you divide by k+1, then it is even smaller. (And x=-1 is included in convergence, because it is alternating harmonic series when x=-1)

17. SolomonZelman

So it should really be: $$-1 \le x<1$$