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Using the fact that $$2^n=\sum^n_{k=0}\binom{n}{k}$$, we can generalize this sum and say that $$2^n=1+n+\frac{n(n-1)}{2!}+\frac{n(n-1)(n-2)}{3!} +...$$such that $n \in \Bbb Z, n \ge0$

Now notice how the constant in the last factor of each term is $1$ less than the number we're taking the factorial of in the denominator. As a result, we subtract $1$ from both sides and divide by $n$. We get $$\frac{2^n-1}{n}=1+\frac{(n-1)}{2!}+\frac{(n-1)(n-2)}{3!}+...$$ this allows us to take the limit as $n$ approaches $0$ while keeping the RHS intact. We get $$\lim_{n\to0}\frac{2^n-1}{n} =1-\frac{1}{2}+\frac{1}{3}+...$$ this limit is easily solved using L'Hôpital's rule and the limit evaluates to $\ln2$.

Is this proof sound? Or did I just get lucky somewhere along the way? Thank you in advance.

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  • $\begingroup$ @IskyMathews: the upper limit is implicit in the binomial (for $n$ positive integer) $\endgroup$
    – G Cab
    Jun 1, 2018 at 11:02
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    $\begingroup$ @IskyMathews It's Newton's generalized binomial theorem. It follows directly from the Taylor expansion of $(1+x)^\alpha$, where $\alpha\in \mathbb{C}$. Convergence in $x=1$ is secured by the fact that $\alpha$ is a positive real number. $\binom{n}{k}=0$ whenever $k>n$ by the (integer) definition. My only doubts about his finding is that the RHS is a series, so we need to swap $n\to 0$ and the infinite sum in some way. $\endgroup$ Jun 1, 2018 at 11:09
  • $\begingroup$ You say "we can generalize this sum ... such that $n\in\mathbb{Z}$". But this is the original sum, not a generalization. From the rest, it seems you're taking $n\in\mathbb{R}$; is that what you meant to do in the first sentence as well? $\endgroup$
    – Teepeemm
    Jun 1, 2018 at 14:09
  • $\begingroup$ Yes, this is correct, modulo some rigour. These kinds of things don't just randomly "get lucky". $\endgroup$ Jun 1, 2018 at 14:32
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    $\begingroup$ @AbhimanyuPallaviSudhir - I think you misunderstood what was meant by "get lucky". It is actually quite common for students to arrive at the correct answer when solving problems by following paths filled with errors. Sometimes because they know what the answer should be and just keep searching till they stumble on it. (My father had a "teacher" who did this, even including the problem number in the calculation, if it would get her to the answer given in her book.) Other times simply by chance. This is what the OP was worried about. $\endgroup$ Jun 1, 2018 at 16:46

3 Answers 3

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All the matter stems from that , for real $r$, the expression $$ \left( {1 + x} \right)^{\,r} = \sum\limits_{0\, \le \,k} { \binom{r}{k} x^{\,k} } = \sum\limits_{0\, \le \,k} {{{r^{\,\underline {\,k\,} } } \over {k!}}x^{\,k} } \quad \left| {\;r \in \mathbb R} \right. $$ (where we indicate with $r^{\,\underline {\,k\,} }$ and $r^{\,\overline {\,k\,} }$, respectively, the Falling and Rising Factorial)

  • converges absolutely , for whichever $r$, if $|x|<1$;
  • for $x=1$ , it converges for $-1<r$;
  • for $0 \le r \in \mathbb Z$ the sum is finite, and thus converges absolutely, for whichever $x$.
    re. to this article in Wikipedia.
    Note in fact that if $r$ is not a non-negative integer, the sum will contain infinitely many terms with alternated sign.

We know that we are allowed to drift inside the sum some algebraic manipulation (including taking the limit) if the sum converges absolutely, while if the sum is just convergent then the convergence might be compromised.
Therefore let's proceed cautiously

For $|x|<1$ we can write $$ \eqalign{ & {{\left( {1 + x} \right)^{\,r} - 1} \over r} = \sum\limits_{1\, \le \,k} {{{r^{\,\underline {\,k\,} } } \over {r\,k!}}x^{\,k} } = \sum\limits_{0\, \le \,k} {{{r^{\,\underline {\,k + 1\,} } } \over {r\,\left( {k + 1} \right)!}}x^{\,k + 1} } = \cr & = \sum\limits_{0\, \le \,k} {{{\left( {r - 1} \right)^{\,\underline {\,k\,} } } \over {\left( {k + 1} \right)!}}x^{\,k + 1} } = x\sum\limits_{0\, \le \,k} {{1 \over {k + 1}}{{\left( {r - 1} \right)^{\,\underline {\,k\,} } } \over {k!}}x^{\,k} } = \int_0^x {\left( {1 + t} \right)^{\,r - 1} dt} \cr} $$ and the integral indicates that, for $r$ approaching $0$, we are running over the edge $$ \int {t^{\,r - 1} dt} = {1 \over r}t^{\,r} \quad \int {t^{\, - 1} dt} = \ln (t) $$ At the same time, the integral is well defined for $x \to 1^{-}$ and for $r \to 0$.

So $$ \eqalign{ & \mathop {\lim }\limits_{r\; \to \,0} {{\left( {1 + x} \right)^{\,r} - 1} \over r} = \ln \left( {1 + x} \right) = \sum\limits_{0\, \le \,k} {{{\left( { - 1} \right)^{\,\underline {\,k\,} } } \over {\left( {k + 1} \right)!}}x^{\,k + 1} } = \cr & = \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} {{1^{\,\overline {\,k\,} } } \over {\left( {k + 1} \right)!}}x^{\,k + 1} } = \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} {{k!} \over {\left( {k + 1} \right)!}}x^{\,k + 1} } = \cr & = \sum\limits_{0\, \le \,k} {{{\left( { - 1} \right)^{\,k} } \over {\left( {k + 1} \right)}}x^{\,k + 1} } \cr} $$ which is the well known Mercator series, and known to be convergent for $-1<x \le 1$.
Note: above we made use of the fact that, for whichever integer $k$ and real (or complex) $s$ we have $$ \eqalign{ & \left( { - s} \right)^{\,\underline {\,k\,} } = \left( { - s} \right)\left( { - s - 1} \right) \cdots \left( { - s - \left( {k - 1} \right)} \right) = \cr & = \left( { - 1} \right)^{\,k} s\left( {s + 1} \right) \cdots \left( {s + \left( {k - 1} \right)} \right) = \left( { - 1} \right)^{\,k} s^{\,\overline {\,k\,} } \cr} $$

So we can take the limit for $x \to 1^{-}$, and obtain $$ \mathop {\lim }\limits_{x\; \to \,1^{\, - } } \ln \left( {1 + x} \right) = \ln 2 = \sum\limits_{0\, \le \,k} {{{\left( { - 1} \right)^{\,k} } \over {\left( {k + 1} \right)}}} $$

There are plenty of posts herewith dealing with this sum, but refer in particular to this post to understand how "delicate" it is: you cannot rearrange the terms (for instance).

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    $\begingroup$ It might be best to mention explicitly that the notation means $$r^{\,\underline {\,k\,}} := r(r-1)(r-2)...(r-k+1)$$ as this is not a widespread standard. And it is not clear to me yet what $1^{\overline k}$ means. $\endgroup$ Jun 1, 2018 at 17:02
  • $\begingroup$ @PaulSinclair: you are right: it is always better to clarify the symbols used. I added to my answer. Thanks for signalling. In any case your intuition for the "falling factorial" is right, while the other is just the "rising" counterpart, so that $1^{\overline k}$ is just $1\cdot2\cdot \cdots \cdot (1+k-1)=k!$ $\endgroup$
    – G Cab
    Jun 1, 2018 at 17:37
  • $\begingroup$ This post is extremely useful and informative, thank you! $\endgroup$
    – Badr B
    Jun 2, 2018 at 1:30
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    $\begingroup$ @BadrB: glad that you appreciate: I saw much confusion around this subject, so I tried my best to set down some pins. $\endgroup$
    – G Cab
    Jun 2, 2018 at 12:04
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To expand on the other answers, you are missing the argument that this also holds for $n>0$ real (else you cannot just take the real limit to $0$). This can be solved by using the binomial series.

$$2^n=1+n+\frac{n(n-1)}{2!}+\frac{n(n-1)(n-2)}{3!} +...$$ is indeed true for any $n\in\mathbb{Z}_{>0}$. Looking at the Maclaurin series of $$(1+x)^n$$ with $n>0$ given by $$1+nx+\frac{n(n-1)x^2}{2!}+\ldots$$ which is convergent in $x=1$, since $\text{Re}(n)>-1$. Now we get exactly $$2^n=1+n+\frac{n(n-1)}{2!}+\frac{n(n-1)(n-2)}{3!} +...$$ for all $n>0$. Indeed, this implies $$\frac{2^n-1}{n}=1+\frac{(n-1)}{2!}+\frac{(n-1)(n-2)}{3!}+...:=\sum_{k=1}^\infty f_n(k)$$ for $0<n<1$ (we choose this interval because then the sum will be infinite: only for $n$ integer it's finite).

Now the big problem is that we need to swap a limit and an infinite sum on the RHS. If we have this, then indeed $$\ln(2)=\lim_{n\downarrow 0}\frac{2^n-1}{n} =\lim_{n\downarrow 0}\sum_{k=1}^\infty f_n(k)=\sum_{k=1}^\infty \lim_{n\downarrow 0}f_n(k)=1-\frac{1}{2}+\frac{1}{3}+...$$ However, this cannot be solved by the dominated convergence theorem: we need to find some function $g:\mathbb{N}\to\mathbb{R}$ such that $|f_n(k)|\leq g(k)$ for all $k\in\mathbb{Z}>0$ and $\sum_{k=1}^\infty g(k)$. This means that $g(k)$ will at least be the harmonic series, which does not converge.

Assuming that $1-\frac{1}{2}+\frac{1}{3}+...$ converges, we have pointwise convergence on compact interval $[0,\frac{1}{2}]$. Dini's theorem doesn't work either, since the series is alternating, so not increasing/decreasing.

Trying to prove uniform convergence also requires estimates with absolute values, resulting in the harmonic series.

With these tools it doesn't work, so this must be such a case where something is true, but the tools break when trying them out on your specific proof. I conclude that you just got lucky somewhere along the way.

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  • $\begingroup$ Thank you. You do raise a good point about convergence. I suppose this proof isn't the best if we don't know whether the series converges or diverges. Other proofs of convergence mentioned by others seem to be using concepts that themselves would yield a "better" proof (i.e. the Mercator series). $\endgroup$
    – Badr B
    Jun 2, 2018 at 3:09
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The equality$$2^n=\sum_{k=0}^n\binom nk$$holds when $n\in\mathbb{Z}_+$. It doesn't make sense to talk about$$\lim_{n\to0}\frac{2^n-1}n$$using it. Besides, how did your factorials vanished?

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    $\begingroup$ How about the binomial series in $x=1$? Factorials vanish for $n$ integer because $\frac{n(n-1)(n-2)\cdots(n-k)}{k!}=0$ when $k>n$. $\endgroup$ Jun 1, 2018 at 11:27
  • $\begingroup$ Or if you mean why the factorials in e.g. $\frac{(n-1)(n-2)}{3!}$ vanish, it's after taking the limit $n\to 0$, so it turns into $\frac{1}{3}$. I think the big question is why we can swap limit and infinite sum. $\endgroup$ Jun 1, 2018 at 11:41
  • $\begingroup$ @ThePhenotype This last possibility is what I meant. $\endgroup$ Jun 1, 2018 at 13:06
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    $\begingroup$ This is not correct. You can use Newton's binomial theorem, and you'll get the same result. In fact, this is implied in the question, since the series on the right is an infinite series. $\endgroup$ Jun 1, 2018 at 14:24
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    $\begingroup$ That what converges? The binomial series clearly converges for all $n$ close to 0 when the base is 2. $\endgroup$ Jun 1, 2018 at 18:21

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