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let $<a_n>$ be a sequence of real numbers such that $\sum|a_n-a_{n-1}|$ is convergent series. Show that power series $\sum a_n x^n$ converge on interval $(-1,1)$

How to approach . let $0<\alpha <1 $. Then i need to show series $\sum a_n \alpha^n$ is convergent. i am trying to show its sequence of partial sums $<S_n>$ is cauchy sequence . $$|S_n-S_m| \leq |a_{m+1}|+|a_{m+2}|+....+|a_n|$$ for $n\geq m$

But then how to use given convergent series? Any hint

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$|a_n| \leq |a_1|+|a_2-a_1|+|a_2-a_3|+\cdots+|a_n-a_{n-1}|$. From this and the hypothesis conclude that $(a_n)$ is bounded. If $|a_n| \leq M$ then $|a_n x^{n}| \leq M|x|^{n}$ and $\sum M|x|^{n}$ is a convergent geometric series. By Comparison Test the series $\sum a_nx^{n}$ is absolutely convergent.

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In Apostol's Mathematical Analysis at page no. 213 exercise 8.27 states $\sum x_ny_n$ converges if $\sum x_n$ converges and $\sum (y_n-y_{n+1})$ converges absolutely.

Now in your problem take $x_n=\alpha^n$ and $y_n=a_n$. For $\alpha\in(-1,1)$, $\sum \alpha^n$ converges and according to the given condition $\sum (a_n-a_{n+1})$ converges absolutely.

This exercise in Apostol can be solved using similar techniques used in the proofs of Dirichlet's criterion and Abel's criterion.

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theres the Abel Theorem which says that if the series converges in two points then converges in the interval (check it out for more details)

in case you dont want to use Abel's Theorem:

you can also use Leibniz to conclude for - 1

for 1 is easy and for any x in between you can use the geometric series + abel dirichlet to conclude

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