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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!
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!

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freckles
 one year ago
Best ResponseYou've already chosen the best response.0Can you show me your series?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0Sometimes it is pretty easy with derivative. Sometimes it is easy with algebra. like we can show \[a_{n+1}<a_n \text{ for }n \text{ using algebra } \\ \text{ or show } f' \text{ is negative for all } x\]

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0could you tell me all the tests? I am trying to review my calc 2 to be ready for calc 3

freckles
 one year ago
Best ResponseYou've already chosen the best response.0Those are the only two I can recall.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0In alternating series, if the absolute value of the terms decreases then the series will converge.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0So, I will denote the sums with \(\rm S_1,~S_2...\) dw:1441594022749:dw

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0that is what i have so far

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Can you post it in here, more clearly.....

freckles
 one year ago
Best ResponseYou've already chosen the best response.0Is the attachment bringing up a new question?

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0no just showing my work so far

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[k+1>k \text{ so } \frac{1}{k+1}<\frac{1}{k}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.0or. ... \[f(k)=\frac{1}{k+1} \\ f'(k)=\frac{1}{(k+1)^2} \\ \frac{1}{(k+1)^2}<0 \text{ for all } k \text{ except } k=1 \\ \text{ which we aren't concerned about }\] either one of these ways shows a_k is a decreasing sequence

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0\(\large\color{black}{ \displaystyle \sum_{ k=0 }^{ \infty } \frac{x^k}{k+1}}\) \(\large\color{black}{ \displaystyle \lim_{ k\rightarrow \infty }\frac{x^{k+1}(k+1)}{(k+2)x^k}}\) \(\large\color{black}{ \displaystyle x\lim_{ k\rightarrow \infty }\frac{k+1}{k+2}}\) \(\large\color{black}{ \displaystylex <1 }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0and 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)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0So it should really be: \(1 \le x<1\)
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