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Radius of convergence of

$$\sum_{n=2}^{\infty} \frac{n^{2n}}{4^n(2n+1)!} (3-2x)^n$$

The answer was to let $$c_n = \frac{(n^{2}(-2))^n}{4^n(2n+1)!}$$

But how do I get the $-2$? Its probably from $(3-2x)^n$. But how do I remove the $3$ and $x$ parts?

Also later into the answer, $$(\lim_{n\to\infty} ... (\lim_{n\to\infty} (1+\frac{1}{n})^n)^2) = \frac{e^2}{8}$$

What gives $e$ btw?

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2 Answers 2

up vote 2 down vote accepted

Factor out a $-2$ from $(3-2x) $. This gives $$\textstyle(3-2x)=\bigl( (-2)\cdot {-3\over2}+ (-2) \cdot x\bigr)=(-2)(x-{3\over2});$$ so then $(3-2x)^n= (-2)^n(x-{3\over2} )^n$.

Recall that a power series has the form $\sum\limits_{n=1}^\infty c_n (x-a)^n$; so the $c_n$ for your series is as you've written (that your series starts at $n=2$ doesn't matter).

As for your second question, you should know that $\lim\limits_{n\rightarrow\infty}\bigl(1+{1\over n}\bigr)^n=e$. This is a fundamental limit.

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For the 2nd qn. Does the values for $1$ matter? eg. If I have $\lim_{n\to\infty} (2+\frac{1}{n})^n$ or $\lim_{n\to\infty} (2+\frac{2}{n})^n$ etc? Or must it be always 1? For the 2nd one, I think its $2e$ right? –  Jiew Meng Mar 25 '12 at 2:33
    
@JiewMeng The more general formula is $\lim\limits_{n\rightarrow\infty}\bigl(1+{x\over n}\bigr)^n=e^x$. The expression $\bigl(2+{1\over n}\bigr)^n$ would tend to infinity as $n$ tends to infinity (you have a number greater than 2 raised to larger and larger powers). So, the "1"'s matter... –  David Mitra Mar 25 '12 at 2:37

I've seen something similar around. Since you need the $x$ part to be of the form $(x-a)^n$, you have to extract the $2$ in the $x$:

$$(3-2x)^n=(-2)^n \cdot (x-3/2)^n$$

Now you have

$$c_n = \frac{(-2n^2)^n}{4^n(2n+1)!}$$

$$c_n = \frac{(-2)^nn^{2n}}{4^n(2n+1)!}$$

You now need to evaluate the limit:

$$\lim\limits_{n \to \infty} \left| \frac{(-2)^{n+1}{(n+1)}^{2(n+1)}}{(-2)^nn^{2n}}\frac{4^n(2n+1)!}{4^{n+1}(2n+3)!}\right|=$$

$$\lim\limits_{n \to \infty}\left| \frac{(-2){(n+1)}^{2n}{(n+1)}^2}{n^{2n}}\frac{1}{4(2n+2)(2n+3)}\right|=$$

$$\frac{1}{2}\lim\limits_{n \to \infty} \left|{\left(\frac{{n+1}}{n}\right)^n}^2\frac{n^2+2n+1}{4n^2+10n+6}\right|=$$

$$\frac{1}{2}\lim\limits_{n \to \infty}\left| {\left(1+\frac{{1}}{n}\right)^n}^2\frac{n^2+2n+1}{4n^2+10n+6}\right|=$$

This limits shouldn't be hard for you, I guess.

$$\displaystyle \frac{1}{2}e^2\frac{1}{4}=\frac{e^2}{8}$$

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