The infinite geometric series \( \displaystyle\sum_{r=0}^\infty \frac9{(y^2-4y+5)^{r-1}} \) where \(y=x^2-6x+11\) has a finite sum then \(x\) cannot be equal to:

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TopNewestRewrite \((y^2 - 4y + 5) \) as \(\displaystyle( (y-2)^2 + 1)\)

Now substitute the value of \(y\) in terms of \(x\) to get(I am just writing the denominator part)

\(\displaystyle ((x^2 - 6x + 9)^2 + 1)^{r-1}\)

Observe that the quadratic of \(x\) is a perfect square of \((x-3)\) hence the expression reduces to

\(\displaystyle ((x-3)^4 + 1)^{r-1}\)

Since the sum should converge the common ratio of the G.P should lie between -1 and 1 And clearly sum will not converge when \(x=3\) because we will get \(9+9+9+9 + \ldots \) infinite times. – Krishna Sharma · 1 year, 3 months ago

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@Krishna Sharma ,thanks a lot – Radhesh Sarma · 1 year, 3 months ago

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