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Answer:
See convergence analysis below
Step-by-step explanation:
For the expansion of 1/(1-t) in powers of t, we can recall that an infinite Geometric series with first term "1" and common ratio "t", in the case that the absolute value of "t" is smaller than 1, can be represented by:
[tex]\Sigma\limits^\infty_0\, t^n=\frac{1}{1-t} \\[/tex]
The condition for such, is that the absolute value of "t" is smaller than "1", which gives us the radius of convergence for this series "1" and the interval of convergence: - 1 < t < 1
Recall as well that using the geometric sequence [tex]1 + t + t^2 + t^3 +...[/tex]
The ratio test becomes [tex]\frac{a_{n+1}}{a_n} = t[/tex]
which is the SAME as the common ratio for the geometric sequence, and which in order to converge to the value [tex]\frac{1}{1-t}[/tex] requires that the absolute valu of the common ration be smaller than "1" rendering the same radius of convergence as discussed above.