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Let $\sum a_n$ be a convergent series of numbers. and let $f: \mathbb R\rightarrow \mathbb R $, be a continuous funtion. Prove that:

$$ \sum [f(a_{n+1})-f(a_n)]$$ is a convergent series.

give an example for that if we don't have a continuous function this this isn't necessarily true.

SOLUTION ATTEMPT: I thought about the definition of continuity , and then linking this to the integral test, and also Heine Theorem that links between Sequences and Functions. but I'm not sure if this is the right direction. any kind of help would be appreciated!

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    $\begingroup$ For the counter example: take $$ f\colon x\mapsto \begin{cases} 1 & \text{ if } x \geq 0\\-1&\text{ otherwise.}\end{cases}$$ and $a_n=\frac{(-1)^n}{n}$. $\endgroup$ – Clement C. Jan 15 '16 at 15:38
  • $\begingroup$ Also, note that you can relax the assumption -- continuity of $f$ at $0$ seems to be sufficient. $\endgroup$ – Clement C. Jan 15 '16 at 18:02
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You cannot prove this with a test that investigates absolute convergence because it is not guaranteed. Rather, write out the $n$-th partial sum explicitly and prove that it converges as $n\to\infty.$

For a counterexample, let $a_n$ be any sequence that is convergent but not absolutely convergent (such as the alternating harmonic sequence in Clement C.'s comment to your question) and let $f$ be the sign function.

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  • $\begingroup$ I got that $\sum [f(a_{n+1})-f(a_n)] = f(a_{n+1})-f(0)$, now i can prove that $f(a_{n+1})-f(0)$ converges to zero by Hiene Theorem plus continuity of $f$ and show that the n -th partial sum sequence converges to final limit then the whole series converges? $\endgroup$ – F1sargyan Jan 15 '16 at 16:11
  • $\begingroup$ If the Heine theorem is the one that says continuous functions between metric spaces preserve limits: then, yes. And convergence of the sequence of partial sums is the definition of convergence of a series. $\endgroup$ – Justpassingby Jan 15 '16 at 16:17
  • $\begingroup$ does it have to converge to $0$ ? $\endgroup$ – F1sargyan Jan 15 '16 at 16:46
  • $\begingroup$ No, to any finite limit and that limit is then called the sum of the series. Now can you write an expression for this limit in our particular case? $\endgroup$ – Justpassingby Jan 15 '16 at 19:37
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Let $S_{m}= \sum_{n=1}^{m} [f(a_{n+1})-f(a_n)]$ be the m-th partial sum and let $L$ be $\lim_{m \rightarrow \infty}{a_{m}}$ . Notice that $S_{m}=f(a_{m+1})-f(a_1)$.

Now, $$\lim_{m \rightarrow \infty}{S_{m}} = \lim_{m \rightarrow \infty}{[f(a_{m+1})-f(a_1)]}=[\lim_{m \rightarrow \infty}{f(a_{m+1})]}-f(a_1)=f(\lim_{m \rightarrow \infty}{a_{m+1})}-f(a_1)=f(L)-f(a_1)$$

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