Let $(a_n)_{n=0}^{\infty}$ be a sequence of real numbers satisfying $\sum_{1}^{\infty} |a_n - a_{n-1}| < \infty$

Which of the following conclusions must then be true?

The series $\sum a_nx^n$ converges

  1. Nowhere on $ \mathbb{R}$
  2. Everywhere on $ \mathbb{R}$
  3. On some interval containing $(-1,1)$
  4. Only on $(-1,1)$

What I have tried is Make an inequality with $|a -b| \leq |a| +|b|$ .. but i cant proceed further .. Any sort of help will be deeply appreciable .

  • 2
    $\begingroup$ Hint. You have $$(1-x)\sum_{n=0}^{\infty} a_n x^n = a_0 + \sum_{n=1}^{\infty} (a_n - a_{n-1}) x^n $$ whenever the series $\sum_{n=0}^{\infty} a_n x^n$ converges. $\endgroup$ Commented Sep 12, 2017 at 18:51
  • $\begingroup$ How to proceed further with this $\endgroup$
    – user469463
    Commented Sep 13, 2017 at 12:06

1 Answer 1


Let $K=\sum_{n=1}^{\infty}|a_n-a_{n-1}|.$ For $n\geq 1$ we have $$|a_n|\leq |a_0|+|a_n-a_0|= |a_0|+|\sum_{j=0}^{n-1}(a_j-a_{j+1})|\leq |a_0|+\sum_{j=0}^{n-1}|a_j-a_{j+1}| \leq |a_0|+K.$$

So $\sup \{|a_n|:n\geq 0\}\leq |a_0|+K<\infty.$

Therefore $\sum_{n=0}^{\infty} a_nx^n $ converges whenever $|x|<1,$ by comparison to the (absolutely convergent) geometric series $\sum_{n=0}^{\infty}(|a_0|+K)|x|^n.$

The correct answer is #3. Case #4 is false if $a_n=0$ for every $n$ .(Case #4 is also false if $a_n=1/n!$ for every $n$, when the power series converges for all $x.$) Case #2 is false if $a_n=1$ for every $n,$ and $x=1$. (Case #2 is also false if $a_n=1+\frac {1}{n+1}$ for every $n,$ and $|x|\geq 1.$)


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