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I was hoping someone could help me with this interesting situation that came up while I was teaching intervals of convergence today using the ratio test.

The problem asked to find the radius and interval of convergence of the series: $$ \displaystyle\sum_{n=1}^{\infty}\,\displaystyle\frac{(-1)^n\cdot x^{2n-1}}{(2n-1)!} $$

Naturally, this is the Taylor series for $-\sin(x)$, so it must converge for all values of $x\in\mathbb{R}$. I then wanted to show how the ratio test could still be used for this, so we did the following: $$\displaystyle\lim_{n\rightarrow\infty}\left|\displaystyle\frac{a_{n+1}}{a_n}\right|=\displaystyle\lim_{n\rightarrow\infty}\left|\displaystyle\frac{(-1)^{n+1}\cdot x^{2(n+1)-1}}{(2(n+1)-1)!}\cdot \displaystyle\frac{(2n-1)!}{(-1)^{n}\cdot x^{2n-1}}\right|\leq 1$$ Now, when we simplify this, we can cancel the negatives and will have: $$\displaystyle\lim_{n\rightarrow\infty}\left|\displaystyle\frac{x^2}{(2n+1)\cdot 2n}\right|\leq1$$ Now, at this point, I would have naturally just factored out the $x^2$ since it is independent of $n$, leaving: $$x^2\cdot\displaystyle\lim_{n\rightarrow\infty}\left|\displaystyle\frac{1}{(2n+1)\cdot 2n}\right|\leq1 \quad \quad \Rightarrow \quad \quad x^2\cdot 0 \leq 1 \quad \quad \Rightarrow \quad \quad x\in\mathbb{R}$$ But, a student asked me why is it that we do not have to worry about the size of $x$ relative to the limit of 0. I had recently taught them L'Hopital's Rule and emphasized the indeterminate form $0\cdot \infty$, and how at times the growth to $\infty$ trumps the infinitesimal growth to 0 in the product, so such a product might head to $\infty$. A simple example is $\displaystyle\lim_{x\rightarrow\infty} 2^x \cdot \frac{1}{x}=\infty$, not 0.

My question is: Why do we not worry about the size of $x$ relative to the limit at $\infty$ of the remaining terms with $n$ in such a problem? I know that $x$ is not in a limit to $\infty$, but is it not possible that for a large enough function of $x$, say $2^x$, that this product will exceed 1, and hence giving a radius of convergence of 0 instead of $\infty$?

Thank you for your help.

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It's not about the 'size' of $x$ and $n$, it's about how they grow in relation to each other.

If $x$ is independent of $n$, the convergence of the limit as $n\to\infty$ will also be independent of $x$.

Even if $x$ was larger than we can comprehend, but still finite, it will never be 'large enough' to compete with $\lim_{n\to\infty}\frac1n$ since this goes to $0$ no matter what $x$ is set to.

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  • $\begingroup$ Thank you for your help with this. I appreciate the prompt and useful feedback. $\endgroup$ – Mada H Mar 24 '15 at 22:28
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$x$ is in $\mathbb R$, so $x < \infty$. It is not allowed to approach $\infty$. No matter how large $x$ is, we do not need to worry about it.

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  • $\begingroup$ Thank you for your help with this. I appreciate the prompt and useful feedback. $\endgroup$ – Mada H Mar 24 '15 at 22:28

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