infinite series - 4

logistic_guy

logistic_guy

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\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle \sum_{k=1}^{\infty}(-1)^{k+1}\frac{3}{k}\)
 
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We will check if \(\displaystyle a_{k+1} \leq a_k\) holds.

\(\displaystyle \frac{3}{k+1}\leq \frac{3}{k}\)


\(\displaystyle \frac{k}{k+1}\leq 1\)

Since this is true for all \(\displaystyle k \geq 1\), then

\(\displaystyle \sum_{k=1}^{\infty}(-1)^{k+1}\frac{3}{k}\) \(\displaystyle \textcolor{blue}{\text{converges.}}\)
 
This criterion doesn't apply to alternating series. By that argument, [imath] \sum k^{-1} [/imath] would converge; however, it diverges.
 
fresh_42 said:
This criterion doesn't apply to alternating series. By that argument, [imath] \sum k^{-1} [/imath] would converge; however, it diverges.
Click to expand...
It is a Theorem \(\displaystyle \rightarrow\) Alternating Series Test.
 
logistic_guy said:
It is a Theorem \(\displaystyle \rightarrow\) Alternating Series Test.
Click to expand...
... which you did not mention. You only said [imath] |a_{n+1}|<|a_n| .[/imath] This is an example of a half-truth. Spitting out something and neglecting the essential part plus the second condition [imath] \displaystyle{\lim_{n \to \infty}a_n=0}. [/imath]

And before I do the same:
... which also mentions the important rearrangement theorem.
 
fresh_42 said:
... which you did not mention. You only said [imath] |a_{n+1}|<|a_n| .[/imath] This is an example of a half-truth. Spitting out something and neglecting the essential part plus the second condition [imath] \displaystyle{\lim_{n \to \infty}a_n=0}. [/imath]

And before I do the same:
... which also mentions the important rearrangement theorem.
Click to expand...
The rest parts of the theorem are trivial. By the way, I am an expert in infinite series. I just solve them to entertain myself and the audience!

😉
 
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