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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.

Badr B
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  • @IskyMathews: the upper limit is implicit in the binomial (for $n$ positive integer) – G Cab Jun 01 '18 at 11:02
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    @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. – The Phenotype Jun 01 '18 at 11:09
  • 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? – Teepeemm Jun 01 '18 at 14:09
  • Yes, this is correct, modulo some rigour. These kinds of things don't just randomly "get lucky". – Abhimanyu Pallavi Sudhir Jun 01 '18 at 14:32
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    @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. – Paul Sinclair Jun 01 '18 at 16:46
  • I agree in general, but the proof in the question is quite clearly straightforward and correct -- showing that this is true is just a question of some degree of formalisation. – Abhimanyu Pallavi Sudhir Jun 01 '18 at 18:17
  • @Teepeemm I was meaning extending the number of terms from $n$ to $\infty$, because we're now using the formula for $\binom{n}{k}$ where $k$ can be greater than $n$. – Badr B Jun 02 '18 at 01:22

3 Answers3

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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).

G Cab
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    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. – Paul Sinclair Jun 01 '18 at 17:02
  • @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 $1\cdot2\cdot \cdots \cdot (1+k-1)=k!$ – G Cab Jun 01 '18 at 17:37
  • This post is extremely useful and informative, thank you! – Badr B Jun 02 '18 at 01:30
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    @BadrB: glad that you appreciate: I saw much confusion around this subject, so I tried my best to set down some pins. – G Cab Jun 02 '18 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.

The Phenotype
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  • 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). – Badr B Jun 02 '18 at 03:09
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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?

José Carlos Santos
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    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 $k>n$. – The Phenotype Jun 01 '18 at 11:27
  • 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. – The Phenotype Jun 01 '18 at 11:41
  • @ThePhenotype This last possibility is what I meant. – José Carlos Santos Jun 01 '18 at 13:06
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    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. – Abhimanyu Pallavi Sudhir Jun 01 '18 at 14:24
  • @AbhimanyuPallaviSudhir And how do you justify that it converges? – José Carlos Santos Jun 01 '18 at 17:13
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    That what converges? The binomial series clearly converges for all $n$ close to 0 when the base is 2. – Abhimanyu Pallavi Sudhir Jun 01 '18 at 18:21
  • @AbhimanyuPallaviSudhir We are talking aboute the series$$1+\binom n1+\binom n2+\binom n3+\cdots$$here. Again, why do you say that it converges? The word “clearly” has no magical properties, as far as I know. – José Carlos Santos Jun 01 '18 at 19:43
  • You bring up a great point. The weakness of this proof is that it provides no concise evidence of convergence. I just assumed that we all agreed that it indeed does converge. My main focus was just finding the sum. – Badr B Jun 02 '18 at 03:12
  • @JoséCarlosSantos Yes, this is the binomial series for $(1+1)^n$. It becomes alternating after a finite number of terms (once $k > n$), and the absolute value is then decreasing and approaches 0, so it converges. This is fairly trivial -- "trivial" has no magical properties, but true nonetheless. – Abhimanyu Pallavi Sudhir Jun 02 '18 at 04:50