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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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!

Mathematics
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Can you show me your series?
Sometimes it is pretty easy with derivative. Sometimes it is easy with algebra. like we can show \[a_{n+1}
n not x

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could you tell me all the tests? I am trying to review my calc 2 to be ready for calc 3
Those are the only two I can recall.
In alternating series, if the absolute value of the terms decreases then the series will converge.
1 Attachment
So, I will denote the sums with \(\rm S_1,~S_2...\) |dw:1441594022749:dw|
that is what i have so far
Can you post it in here, more clearly.....
Is the attachment bringing up a new question?
no just showing my work so far
\[k+1>k \text{ so } \frac{1}{k+1}<\frac{1}{k}\]
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
\(\large\color{black}{ \displaystyle \sum_{ k=0 }^{ \infty } \frac{x^k}{k+1}}\) \(\large\color{black}{ \displaystyle \lim_{ k\rightarrow \infty }\frac{x^{k+1}(k+1)}{(k+2)x^k}}\) \(\large\color{black}{ \displaystyle x\lim_{ k\rightarrow \infty }\frac{k+1}{k+2}}\) \(\large\color{black}{ \displaystyle|x |<1 }\)
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)
So it should really be: \(-1 \le x<1\)

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