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Does $\displaystyle\lim_{x^-\to 2} \sum_{i=1}^\infty (f_x(i)\cdot x)^i$, where $\lim\limits_{i\to \infty} f_x(i) = 1/2$ for $1<x\leq 2$, diverge?

Is the following a valid proof? \begin{eqnarray*} \lim_{x^-\to 2} \sum_{i=1}^\infty (f_x(i)\cdot x)^i &=& \lim_{x^-\to 2} \lim_{n\to \infty}\sum_{i=1}^n(f_x(i)\cdot x)^i\ &=& \lim_{n\to \infty}\lim_{x^-\to 2}\sum_{i=1}^n(f_x(i)\cdot x)^i \ &=& \lim_{n\to \infty}\sum_{i=1}^n(f_2(i)\cdot 2)^i \end{eqnarray*} which diverges because the terms $f_2(i)\cdot 2$ tends to 1 as $i$ tends to infinity. So the question really boils down to wether you can swap the limits.

Edit I have forgotten to add that $f$ also depends on $x$, is linear in $x$, and still converges to $1/2$ as long as $1<x\leq 2$. I have thus changed the limits so that $x$ approches $2$ from the left and added a subscript to $f$.

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Compare your sum to finite sums. Or use Lebesgue theorem for series with nonnegative terms.. –  Did Feb 6 '12 at 19:11
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Limit exchanges have to be done carefully; for example, consider $$\lim_{n\to 0}\sum_{i=1}^{\infty}\frac{n}{i}=\lim_{n\to 0}\lim_{m\to\infty}\sum_{i=1}^m\frac{n}{i}.$$The limit diverges, because for each $n$, $\sum\frac{n}{i}$ diverges; but after exchanging limits we have $$\lim_{m\to\infty}\lim_{n\to 0}\sum_{i=1}^{m}\frac{n}{i}=0$$because for each $m$, the sum goes to $0$. So swapping limits is not always valid. It is valid in some instances, so you need to find a method to justify the swap in this instance, or try something else. –  Arturo Magidin Feb 6 '12 at 19:15

1 Answer 1

up vote 1 down vote accepted

Your answer is correct provided that the limit $\lim_{x^-\to 2}\sum_{i=1}^n(f_x(i)\cdot x)^i$ is uniform with respect to n. The reason for this is the following theorem. And their justification consists in applying the triangle inequality exhaustive.

Theorem. Let $\{ F_t ; t\in T\}$ be a family of functions $F_t : X \rightarrow \mathbb{C}$ depending on a parameter t; let $\mathcal{B}_X$ be a base $X$ and $\mathcal{B}_{T}$ a base in $T$. If the family converges uniformly on $X$ over the base $\mathcal{B}_{T}$ to a function $F : X \rightarrow \mathbb{C}$ and the limit $\lim_{\mathcal{B}_{T}} F_t(x)=A_t$ exists for each $t\in T$, the both repeated limits $\lim_{\mathcal{B}_{X}}(\lim_{\mathcal{B}_{T}}F_t(x))$ and $\lim_{\mathcal{B}_{T}}(\lim_{\mathcal{B}_{X}}F_t(x))$ exist and the equality

$$ \lim_{\mathcal{B}_{X}}(\lim_{\mathcal{B}_{T}}F_t(x))=\lim_{\mathcal{B}_{T}}(\lim_{\mathcal{B}_{X}}F_t(x)) $$ holds.

This theorem can be found in books of Lang (Analysis vol. 2 for example). But I think the wording of Zorich (Mathematical Analysis II p. 381) more generally.

The converse of this theorem is also true.

In this case we have: $T=\mathbb{N}\cup\{\infty\}$, $X=[1,2]$, $\mathcal{B}_{T}=\mathcal{B}_{\mathbb{N}\cup\{\infty\}}$ the set of parts of $\mathbb{N}\cup\{\infty\}$.

and $F_n(x)=\Sigma_{i=1}^n( f(i)\cdot x)^i$. If we prove that $ |F_n(y)-F_n(x)|<L\cdot|y-x|, \quad L>0 $ we get what we wanted. I think using the mean value theorem for $ F_n (x) $ and using appropriate cotes to $ F '(x) $ with $ 2 - \epsilon <x \leq 2$ we can prove that $ L = 1$.

Or if $\lim_{x\to 2}F_n(x)$ is dependent of n so justification is not worth! And the limit does not exist.

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