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Which of the following functions are continuous?

  1. $f(x) = x^2 +x^2/(1 + x^2 )+x^2/(1 + x^2)^2 + \cdots, x \in \mathbb{R}$.
  2. $f(x) =\sum_{n=1}^\infty(−1)^n \cos nx/n ^{3/2} , x \in [−\pi,\pi]$.
  3. $f(x) =\sum_{n=1}^\infty n^2x^n, x \in \left[−\frac12,\frac12\right]$


  1. in $f(x)=1+x^2$ which is obviously continuous.but 1. is not given in the confused.
  2. I guess by alternating series test it is convergent. but how can I find the value of the series.
  3. no idea.

can somebody help me please . thanks for your help.

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In (a), try plugging in $x=0$. – Antonio Vargas Jan 14 '13 at 5:07
For (a) try writing $f$ as a geometric series for $x\neq 0$, for (b) and (c) use Weierstrass' M-test. – Jose27 Jan 14 '13 at 5:10
up vote 1 down vote accepted

For 1. write $f(x) = x^2 \sum_{k=0}^\infty \frac{1}{(1+x^2)^k}$. Evaluate the expression for $x\neq 0$ and $x=0$.

For 2. note that $f(x) = \sum_{k=0}^\infty \phi_k(x) \frac{1}{n^p}$, where $|\phi_k(x)| \leq 1$ for all $x$, and $p = \frac{3}{2} > 1$.

For 3. let $\phi(x) = \sum_{k=0}^\infty x^k$, and note that for $|x| \leq \frac{1}{2}$, the series is uniformly convergent to $\phi(x) = \frac{1}{1-x}$, which is smooth. Hence the summation and differentiation can be interchanged. This gives $\phi'(x) = \sum_{k=0}^\infty (k+1)x^k$ and $\phi''(x) = \sum_{k=0}^\infty (k+1)(k+2)x^k$. Since $k^2 = (k+1)(k+2)-3(k+1)+1$, we have $f(x) = \phi''(x)-3 \phi'(x)+ \phi(x)$. This is sufficient to answer the question, but if you care, you can compute $f$ explicitly as $f(x) = \frac{x(x+1)}{(1-x)^3}$.

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THANKS A LOT.... – poton Jan 14 '13 at 10:04

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