Another exercise from Bartle/Sherbert Introduction to Real Analysis book (this one is exercise 9.4.14):

Use the Lagrange form of the remainder to justify the general Binomial Expansion $$(1+x)^{m}=\sum_{n=0}^{\infty}\binom{m}{n}x^{n}\quad \mathrm{for}\ 0\le x<1$$

Note: $m$ in an arbitrary real number.

My take: To prove the statement, one should show that Taylor series coefficients around $x=0$ are indeed $\frac{f^{(n)}(0)}{n!}=\binom{m}{n}$, that is rather obvious, and then also that the limit of the Lagrange form of the remainder: $$R_{n}(x)=\binom{m}{n+1}(1+c)^{m-(n+1)}x^{n+1}$$ is $0$ when $n\rightarrow\infty$ (of course, it is necessary to show this only for $0\le x<1$ and $0\le c<1$).

I have trouble with this later part. Here, $\frac{x}{1+c}<1$, thus $(\frac{x}{1+c})^{n+1}\rightarrow 0<$ when $n\rightarrow\infty$. Also, $(1+c)^{m}$ won't affect the limit if it's zero. But I'm not sure what to do about $\binom{m}{n+1}$ part...

  • $\begingroup$ For the reference: here is an answer to the same problem, but using Ratio Test to prove the convergence of Taylor series: math.stackexchange.com/questions/124293/…. I'm still interested in proof that the limit of the remainder is $0$ when $n\rightarrow\infty$. $\endgroup$ – Crni Apr 8 '13 at 11:34
  • $\begingroup$ @Cmi Keep in mind that it doesn't suffice to show that the taylor series converges. One must show it converges to $f$. $\endgroup$ – philmcole Jan 27 '18 at 20:24

If $m\in\mathbb{N}$, there is nothing to do. For non-integer $m>0$, let $k\ge0$ be an integer such that $k<m<k+1$. Then $0<m-k<1$, $-1<m-k-1<0$, $-2<m-k-2<-1$ and $n-k-1<n-m<n-k$. Noting that \begin{eqnarray*} \left|\binom{m}{n+1}\right|&=&\left|\frac{m(m-1)\cdots(m-n)}{(n+1)!}\right|\\ &\le&\frac{|m(m-1)\cdots(m-k+1)(m-k)(m-k-1)(m-k-2)\cdots(m-n)|}{(n+1)!}\\ &=&\frac{m(m-1)\cdots(m-k+1)(m-k)|(m-k-1)(m-k-2)\cdots(m-n)|}{(n+1)!}\\ &\le& m(m-1)\cdots(m-k+1)(m-k)\frac{1\cdot 2\cdots (n-k-1)}{(n+1)!}\\ &\to 0 \end{eqnarray*} as $n\to\infty$. Thus \begin{eqnarray*} |R_n(x)|&=&\left|\binom{m}{n+1}(1+c)^{m-(n+1)}x^{n+1}\right|\\ &\le&\left|\binom{m}{n+1}\right|(1+c)^m\left(\frac{1}{1+c}\right)^{n+1}\\ &\le&\left|\binom{m}{n+1}\right|(1+c)^m\\ &\to&0 \end{eqnarray*} For $m<0$, the story is different. We do not have $\left|\binom{m}{n+1}\right|\to 0$ as $n\to\infty$. In fact, let $k\ge 0$ be an integer such that $k\le-m<k+1$. Then \begin{eqnarray*} \left|\binom{m}{n+1}\right|&=&\left|\frac{m(m-1)\cdots(m-n)}{(n+1)!}\right|\\ &=&\left|\frac{-m(-m+1)\cdots(-m+n)}{(n+1)!}\right|\\ &\ge&\frac{(-m)(k+1)\cdots(k+n)}{(n+1)!}\\ &=&\frac{(-m)(n+k)!}{(n+1)!k!}\\ &\to \infty \end{eqnarray*} as $n\to\infty$. Thus $\lim_{n\to\infty}R_n(x)$ diverges.

  • $\begingroup$ I suppose $\binom{n+1}{k}$ and $\binom{k}{n+1}$ are the same which is in your question. $\endgroup$ – xpaul Apr 8 '13 at 16:59
  • $\begingroup$ Thanks, that indeed proves it for $m>0$. I think that, in the fourth row of equations, product in numerator should go up to $n-k$. I removed my previous comments, as they were referring to previous edits of your answer. $\endgroup$ – Crni Apr 8 '13 at 18:27
  • $\begingroup$ For $m=0$, remainder is obviously $0$. And for $m<0$, if $s=|m|$, then $\Big|\binom{m}{n+1}\Big|=\frac{s(s+1)\ldots(s+n)}{(n+1)!}=\prod_{k=0}^{n}\frac{s+k}{k+1}$. Now, on to proving that the limit of this product is somehow bounded, so that from the reasoning I provided in the question, one could conclude that the limit of the whole remainder is $0$... $\endgroup$ – Crni Apr 8 '13 at 18:42
  • $\begingroup$ So would you mind filling in the answer for $m<0$? This is exactly the missing detail that I've put the bounty for; however, as mentioned in my previous comment, I wouldn't agree that this case is same as for $m>0$. $\endgroup$ – Crni Apr 14 '13 at 7:45
  • $\begingroup$ @Crni, I filled the case of $m<0$. It turns out that $\lim_{n\to}R_n(x)$ diverges. $\endgroup$ – xpaul Apr 14 '13 at 16:02

Let me supply the proof that the remainder $R_{n-1}(x)$ tends to $0$ in the case $m<0$.

We can assume $0<c<x<1$ and let $\alpha=\frac{x}{1+c}<1$. It then suffices to show that $\binom{m}{n}\alpha^n=\binom{m}{n}(\frac x {1+c})^n\to 0$. Choose $k\in \mathbb N$ such that $k+1\ge -m$. Then for $\beta=\frac 1{\alpha}$ and $n\ge k$, \begin{align*} \left|\frac{m(m-1)\cdots(m-n+1)}{n!}\alpha^n\right| \le&\frac{(k+1)(k+2)\cdots(k+n)}{n!}\alpha^n\\ =& \frac{1}{k!}\cdot\frac{(k+n)!}{n!}\alpha^n\\ =& \frac{1}{k!}\cdot\frac{(n+1)(n+2)\cdots (n+k)}{\beta^n}\\ \le& \frac{1}{k!}\cdot\frac{(2n)^k}{\beta^n}\\ \to&0 \end{align*} as $n\to \infty$. The last line follows from $\lim_{x\to \infty}\frac{x^k}{\beta^x}=0$, which in turn follows from L'Hospital rule.


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