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Let $f:(-1,1)$ $\to \mathbb{R}$ such that $f^{(n)}(x)$ exists and $|f^{(n)}(x)|\leq 1$ for every $n\geq 1$ and for every $x\in(-1,1)$. Then f has a convergent power series expansion in a neighbourhood of

a) every $x\in(-1,1)$

b) every $x\in(-1/2,0)$ only

c) no $x\in(-1,1)$

d) every $x\in(0,1/2)$ only

I "feel" the answer should be option a). $f^{(n)}(x)$ exists and $|f^{(n)}(x)|\leq 1$ probably points towards to absolute convergence.

This was part of a chapter on infinite series which gave definitions of convergence related concepts (tests of convergence, power series, uniform convergence, etc.) and a few worked out examples. There were no example problems like this one.

What theory covers this type of problem ? Any hints as to how to tackle problems like this ?

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  • $\begingroup$ Your guess is correct! Try to prove it now. $\endgroup$ – science Feb 23 '15 at 4:06
  • $\begingroup$ Could you help me get started towards that ? $\endgroup$ – square_one Feb 23 '15 at 4:11
  • $\begingroup$ You can to apply tests for series convergence! $\endgroup$ – science Feb 23 '15 at 4:30
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Hint: To prove the series converges you could use Wierstrass' M-test. To see that it converges to the function $f$ use an appropriate form of the remainder term (any one will do actually), and prove that $R_n(x)\to 0$ as $n\to \infty$.

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