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I have to show that

$(1-\frac{t^2}{2r}+O(r^{-\frac{3}{2}}))^{-r}\rightarrow e^{\frac{t^2}{2}}$ as $r\rightarrow\infty$

I have expanded the given expression like below:

$(1-\frac{t^2}{2r}+O(r^{-\frac{3}{2}}))^{-r}$

$= (1-\frac{t^2}{2r})^{-r} - rO(r^{-\frac{3}{2}})(1-\frac{t^2}{2r})^{-r-1} + \binom{r+1}{2}(O(r^{-\frac{3}{2}}))^2(1-\frac{t^2}{2r})^{-r-2}-...$

As $r\rightarrow\infty$,

$(1-\frac{t^2}{2r})^{-r}\rightarrow e^{\frac{t^2}{2}}$

$rO(r^{-\frac{3}{2}})\rightarrow 0$, $\binom{r+1}{2}(O(r^{-\frac{3}{2}}))^2\rightarrow 0$ etc.

But what about the terms $(1-\frac{t^2}{2r})^k, k=-r-1, -r-2,....$

How can I say that they tend to a finite value? Can I say they are finite because $(1-\frac{t^2}{2r})^{-r}$ tends to a finite value(i.e. $e^{\frac{t^2}{2}}$)?

user587389
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    Do you know the limit $(1+ \frac1n)^n \to e$? – md2perpe Oct 06 '20 at 21:53
  • @md2perpe Yes. Like here I have written that $(1-\frac{t^2}{2r})^{-r} \to e^{\frac{t^2}{2}}$. What I want to know is whether $(1-\frac{t^2}{2r})^{k} \to$ a finite value($k=-r-1, -r-2$ and so on). I guess these terms do tend to a finite value, but I want someone to clarify the logic behind it. – user587389 Oct 06 '20 at 22:25

1 Answers1

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Using Taylor Theorem, it is easy to show that $\log(1-x)=-x+O(x^2)$, from which we have

$$\begin{align} \left(1-\frac{t^2}{2r}+O\left(r^{-3/2}\right)\right)^{-r}&=e^{-r\log\left(1-\frac{t^2}{2r}+O\left(r^{-3/2}\right)\right)}\\\\ &=e^{t^2/2}e^{-O(r^{-1/2})} \end{align}$$

Therefore, we see that

$$\lim_{r\to \infty }\left(1-\frac{t^2}{2r}+O\left(r^{-3/2}\right)\right)^{-r}=e^{t^2/2}$$

as was to be shown!

EDIT:

The OP was interested in proceeding with a binomial expansion. To that end, we now use a generalized binomial expansion of the term if interest to write

$$\begin{align} \left(1-\frac{t^2}{2r}+O\left(r^{-3/2}\right)\right)^{-r}&=\sum_{k=0}^\infty \binom{-r}{k}\left(1-\frac{t^2}{2r}\right)^{-r-k}O\left(r^{-3/2}\right)^k\tag1 \end{align}$$

where the generalized binomial coefficient, $\binom{-r}{k}$, is given by

$$\binom{-r}{k}=\frac{-r(-r-1)\cdots (-r-k+1)}{k!}$$

Owing to the uniform convergence of the series on the right-hand side of $(1)$, we can interchange the order of the limit and the summation to obtain

$$\begin{align} \lim_{r\to\infty}\sum_{k=0}^\infty \binom{-r}{k}\left(1-\frac{t^2}{2r}\right)^{-r-k}O\left(r^{-3/2}\right)^k&=\sum_{k=0}^\infty \lim_{r\to \infty}\left(\binom{-r}{k}\left(1-\frac{t^2}{2r}\right)^{-r-k}O\left(r^{-3/2}\right)^k\right)\\\\ &=e^{t^2/2} \end{align}$$

since for $k\ne 0$,

$$\binom{-r}{k}O\left(r^{-3/2}\right)^k=O\left(r^{-1/2}\right)^k\to 0$$

as $r\to \infty$ and

$$\lim_{r\to \infty}\left(1-\frac{t^2}{2r}\right)^{-r-k}=e^{t^2/2}$$

Mark Viola
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  • I guess in the last line it will be $e^{t^2}$, not $e^{-t^2}$. And sorry to say, but I cannot get how this answer addresses my issue. What I want to know is whether $(1-\frac{t^2}{2r})^{k} \to$ a finite value($k=-r-1, -r-2$ and so on). I guess these terms do tend to a finite value, but I want someone to clarify the logic behind it. – user587389 Oct 06 '20 at 22:33
  • Thank you for catching the typo. I've edited accordingly. As for your expansion, it seems to have a flaw. – Mark Viola Oct 06 '20 at 22:36
  • @user587389 I've added a new section that addresses your specific question. – Mark Viola Oct 07 '20 at 22:08
  • @user587389 Please let me know how I can improve my answer. I really want to give you the best answer I can. – Mark Viola Oct 10 '20 at 16:18
  • Can you tell me how I can prove the last line of your answer, i.e., $\lim_{r\rightarrow\infty}(1-\frac{t^2}{2r})^{-r-k}=e^{kt^2/2}$? That was my question actually. However, thank you for your effort and sorry for my late reply. – user587389 Oct 21 '20 at 14:54
  • First, you're certainly welcome. It's my pleasure to help. One if the several characterizations of the exponential function is $$e^x\equiv \lim_{n\to \infty}\left(1+\frac xn\right)^n$$Now let $x=-t^2/2$. – Mark Viola Oct 21 '20 at 16:38
  • Oh. There was a typo. There is no $k$ in the exponent. – Mark Viola Oct 21 '20 at 17:13
  • I know that $\lim_{r\rightarrow\infty}(1-\frac{t^2}{2r})^{-r}=e^{\frac{t^2}{2}}$. I want to know if $\lim_{r\rightarrow\infty}(1-\frac{t^2}{2r})^{-r-k}$ will be finite and how?$(k=1,2,...)$ – user587389 Oct 22 '20 at 13:51
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    For each fixed $k$, $\lim_{r\to\infty}\left(1-\frac{t^2}{2r}\right)^{-k}=1$. The uniform convergence property guarantees that we can interchange the limit with the infinite series. Is that clearer? – Mark Viola Oct 22 '20 at 14:01
  • Yes, thanks a lot. – user587389 Oct 22 '20 at 15:54
  • You're welcome. My pleasure. – Mark Viola Oct 22 '20 at 16:49