I need quick help! (1 Viewer)

coyazayo · Jul 9, 2015

Hi I have a question I couldn't really solve. It would be great if you helped me!
Question is about geometric series.

Find the values of x for which the series 1+1/x-1+1/(x-1)squared+... Has a limiting sum.

Apparently you need to consider 2 things when x-1>0 and x-1<0. It's really confusing HELP!

InteGrand · Jul 9, 2015

coyazayo said:
Hi I have a question I couldn't really solve. It would be great if you helped me!
Question is about geometric series.

Find the values of x for which the series 1+1/x-1+1/(x-1)squared+... Has a limiting sum.

Apparently you need to consider 2 things when x-1>0 and x-1<0. It's really confusing HELP!

$Let $S=1+\frac{1}{x-1}+\frac{1}{(x-1)^2}+\cdots$. Clearly the common ratio is $r=\frac{1}{x-1}$. The series has a limiting sum if and only if $|r|<1$, i.e.$

$$-1<\frac{1}{x-1}<1$.$

$One easy way to solve this inequation is just to sketch the graph of $y=\frac{1}{x-1}$, which is just the standard hyperbola $y=\frac{1}{x}$ shifted right by 1 unit. The function's graph has a $y$-intercept of $-1$, and the function equals 1 when $x=2$. The sketch would look like this:$

http://www.wolframalpha.com/input/?i=plot+y=1/(x-1)

$If you sketch it decently enough, you'll see that the range of values of $x$ for which the function is between $-1$ and $+1$ is $x<0$ or $x>2$, which is, therefore, the answer.$

rand_althor · Jul 9, 2015

Alternatively, if you don't want to use a graphical approach:
$\begin{align*}\textup{There is a limiting sum when: }-1&<\frac{1}{x-1}<1 \\\textup{Separate it to }\frac{1}{x-1}&>-1\textup{ and }\frac{1}{x-1}<1 \\\textup{The first inequality:} \\\frac{1}{x-1}&>-1 \\x-1&<-1 \\x&<0 \\\textup{The second inequality: } \\\frac{1}{x-1}&<1 \\x-1&>1 \\x&>2 \\\textup{Therefore the solutions are: }x<0, \: x>2 \end{align*}$

I need quick help! (1 Viewer)

coyazayo

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InteGrand

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rand_althor

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