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We need to prove that $\lim_\limits{x \to \infty} \ln((1 + \frac{1}{x})^x) = \lim_\limits{x \to \infty} x \cdot \ln(1 + \frac{1}{x}) = 1$ and would like to use Taylor series.

To do this, when we expand $\ln(1 + \frac{1}{x})$, we should get a series like $\frac{1}{x} + \frac{a_1}{x^2} + \frac{a_2}{x^3} + ...$, multiplying by $x$ would make it $1 + \frac{a_1}{x} + \frac{a_2}{x^2} + ...$ which is $1$ if $x \to \infty$.

The Taylor expansion of $f(x) = \ln(1 + \frac{1}{x})$ at some $a$ is $\sum_{n = 0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 + \frac{f'''(a)}{6}(x-a)^3 + ...$.

$f'(a) = -\frac{1}{a^2 + a}$, $f'(a) \cdot (x-a) = - \frac{x-a}{a^2 + a}$

$f''(a) = \frac{2a + 1}{(a^2 + a)^2}$, $f''(a) \cdot (x-a) = \frac{(2a + 1)(x-a)}{(a^2 + a)^2}$

So the Taylor series begins like this:

$\ln(1 + \frac{1}{x}) = \ln(1 + \frac{1}{a}) - \frac{x-a}{a^2 + a} + \frac{(2a + 1)(x-a)}{(a^2 + a)^2} - ...$

Since the first term doesn't involve $x$ and we want it to be $\frac{1}{x}$, we would like it to be $0$ and the next term to be $\frac{1}{x}$. To make the first term $0$, $a$ needs to be approaching infinity.

$\lim_\limits{a \to \infty} \left( \ln(1 + \frac{1}{a}) - \frac{x-a}{a^2 + a} + \frac{(2a + 1)(x-a)}{(a^2 + a)^2} - ... \right) = \lim_\limits{a \to \infty} \left(- \frac{x-a}{a^2 + a} + \frac{(2a + 1)(x-a)}{(a^2 + a)^2} - ... \right)$.

Now, this isn't in the form $\frac{1}{x} + \frac{a_1}{x^2} + \frac{a_2}{x^3} + ...$ that we wanted.

Do we need to pick some other $a$?

I found https://math.stackexchange.com/a/1071689/1095885 which says

When x is very large, using Taylor

$\log\left(1+\frac{1}{x}\right)=\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3 x^3}+O\left(\left(\frac{1}{x}\right)^4\right)$

The series $\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{x^3} + ...$, which would be perfect, seems to be a Laurent series though.

Edit: When substituting $u = \frac{1}{x}$ and taking the Taylor series of $\ln(1 + u)$, why don't we have to use the chain rule?

ILikeMathematics
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    I'd substitute $u=1/x$ and work with that – Akiva Weinberger Jan 03 '23 at 19:12
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    @AkivaWeinberger's advice is so good it can spare you Taylor series, because if $f(0)=0$ then$$\lim_{x\to\infty}xf(1/x)=\lim_{u\to0^+}\frac{f(u)}{u}$$is the definition of the right-handed derivative of $f$ at $u=0$, so is the two-sided derivative $f'(0)$ if that exists. – J.G. Jan 03 '23 at 19:29

3 Answers3

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It seems that you're making this a bit more complicated than is necessary.

For $\lvert u \rvert < 1$, $$ \ln(1 + u) = u - \frac12 u^2 + \frac13 u^3 - \frac14 u^4 + \cdots $$ You can verify this Maclaurin series (Taylor series centered at $0$) by differentiating both sides to obtain the well-known geometric series $$ \frac{1}{1+u} = 1 - u + u^2 - u^3 + \cdots $$ and by verifying that the constant term vanishes by evaluating at $u=0$.

Now, by dividing through by $u$, we have the series $$ \frac1u \ln(1 + u) = 1 - \frac12 u + \frac13 u^2 - \frac14 u^3 + \cdots $$ which has the same radius of convergence, using, say the Alternating Series Test.

As $x \to \infty$, we can certainly bound $\lvert x \rvert > 1$, so that $\lvert u \rvert = \lvert \frac1x \rvert < 1$, hence $$ x \cdot \ln\biggl(1 + \frac1x\biggr) = 1 - \frac{1}{2x} + \frac{1}{3x^2} - \frac{1}{4x^3} + \cdots $$ converges too for any such $x$.

It's easy to see that all but the constant term vanish as $x \to \infty$, revealing the limit.

Sammy Black
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  • Thank you. Why is it valid to take the Maclaurin series with respect to $u$? When we get the Maclaurin series of $\ln(1 + u)$, we are taking derivatives. For $\frac{1}{x} + \frac{1}{x^2}$, when we substitute $u = \frac{1}{x}$, we get $u + u^2$. Taking the derivative of that (with respect to $u$), we get $1 + 2u$, but substituting back, that gives us $1 + \frac{2}{x}$, which is not the derivative of $\frac{1}{x} + \frac{1}{x^2}$. Why does the same not happen when we substitute $u = \frac{1}{x}$ and take the Maclaurin series? – ILikeMathematics Jan 03 '23 at 22:00
  • Phrases like "the derivative of ..." are incomplete. Derivative with respect to what quantity? This is what the chain rule is all about! – Sammy Black Jan 03 '23 at 22:29
  • As an example, if $u = \tfrac1x$, then we can calculate $\frac{\mathrm{d}}{\mathrm{d}x} \bigl[ \frac1x + \frac1{x^2} \bigr]$ using the expression in terms of $u$ if we want to: \begin{align} \frac{\mathrm{d}}{\mathrm{d}x} \biggl[ \frac1x + \frac1{x^2} \biggr] &= \frac{\mathrm{d}}{\mathrm{d}x} \biggl[ u + u^2 \biggr] \ &= \frac{\mathrm{d}}{\mathrm{d}u} \biggl[ u + u^2 \biggr] , \frac{\mathrm{d}u}{\mathrm{d}x} \ &= (1 + 2u) , \biggl( - \frac{1}{x^2} \biggr) \ &= \biggl( 1 + \frac{2}{x} \biggr) , \biggl( - \frac{1}{x^2} \biggr) \ &= - \frac{1}{x^2} - \frac{2}{x^3} \end{align} – Sammy Black Jan 03 '23 at 22:29
  • Why do we not have to use the chain rule on the series?

    $$ \ln(1 + u) + \frac{d}{du} \ln(1 + u) \frac{du}{dx} \cdot (x - u) + \frac{d^2}{du^2} \ln(1 + u) \frac{d^2u}{dx^2} \cdot \frac{(x - u)^2}{2} + ...$$

    $$= \ln(1 + u) + \frac{1}{1+u} \cdot (-\frac{1}{x^2}) \cdot (x - u) -\frac{1}{(1 + u)^2} \cdot \frac{2}{x^3} \cdot \frac{(x - u)^2}{2} + ...$$

    For some $u$. For $u = 0$, this is

    $$= -\frac{1}{x} - \frac{1}{x} + ...$$

     – ILikeMathematics Jan 04 '23 at 13:11
  • I think the better way to do this is to start with the series for $1/(1+u)$ and integrate it to get the series for $\ln(1+u)$. – marty cohen Jan 05 '23 at 20:07
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Using Taylor series here is like the proverbial killing a fly with a sledgehammer. Set $h = 1/x$, so the condition $x \to \infty$ is the same as $h \to 0^+$ and $$ x\ln\left(1+\frac{1}{x}\right) = \frac{\ln(1+h)}{h} = \frac{\ln(1+h)-\ln 1}{h}. $$ Therefore $$ \lim_{x \to \infty} x\ln\left(1+\frac{1}{x}\right) = \lim_{h \to 0^+} \frac{\ln(1+h)-\ln 1}{h}, $$ which shows your calculation follows from calculating the derivative of $\ln x$ at $x = 1$ (using $h \to 0$, not just $h \to 0^+$): $$ \lim_{h \to 0} \frac{\ln(1+h)-\ln 1}{h} = \left.\frac{d}{dx}\right|_{x=1}\ln x = \ ?? $$

KCd
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  • Yes, apart from that, l'Hopital can also be used. – ILikeMathematics Jan 06 '23 at 16:24
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    @ILikeMathematics that would be circular reasoning, since the calculation you asked about is equivalent to computing the derivative of $\ln x$ at $x=1$ and using L’Hospital’s rule for your calculation requires knowing the derivative of $\ln x$. – KCd Jan 06 '23 at 16:35
  • We already know the derivative of $\ln(x)$ though, why is it circular reasoning? – ILikeMathematics Jan 06 '23 at 16:47
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    Please tell me how you know the derivative of $\ln x$ is $1/x$. If $f(x)=\log_a x$ for $x > 0$, then properties of logarithms (for any base $a$) and a change of variables in the limit definition of derivatives shows $f’(x)=f’(1)/x$, so $f’(x)=1/x$ if and only if $f’(1)=1$, and that’s equivalent to $\lim_{|y|\to\infty}y\log_a(1+1/y)=1$. So, depending on why you think $\ln’(x)=1/x$, that fact “we already know” is equivalent to the limit you are asking about. It is like using L’Hospital to show $\lim_{x\to 0} (\sin x)/x=1$ for trig functions in radians. – KCd Jan 06 '23 at 19:06
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    I use the term “circular reasoning” because this reasoning uses P to prove Q while forgetting that Q is equivalent to a result R that was already used to prove P. For you, P is $\ln’ x = 1/x$, Q is $\lim_{|x|\to\infty}x\ln(1+1/x)=1$, and R is $\ln’(1)=1$. A more common version of that has P being $\sin’x=\cos x$, Q being $\lim_{x\to 0}(\sin x)/x=0$, and R being $\sin’(0)=1$. – KCd Jan 06 '23 at 19:21
  • Well, you could use another way to arrive at the derivative of $\ln(x)$, for example proof 1 from https://proofwiki.org/wiki/Derivative_of_Natural_Logarithm_Function. – ILikeMathematics Jan 07 '23 at 11:29
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    That proof depends on making a definition of $\ln x$ as a certain integral. While elegant mathematically, it has the idea that $\ln’x = 1/x$ baked into it. And more concretely, I suspect this is surely not how you thought of $\ln x$ before you posted your question. How were you taught that $\ln’x = 1/x$? – KCd Jan 07 '23 at 15:43
  • @ILikeMathematics: I have to agree with the answer as well as comments by KCd. I discussed this issue of using L'Hospital in this answer. – Paramanand Singh Jan 09 '23 at 05:23
  • @ParamanandSingh Thanks. If using L'Hospital is circular in this case depends on how we find $\left(\ln(x)\right)'$, if we find it by $f'(1)/x$, as KCd mentioned, it's circular, if we find it in another way, it's not, right? – ILikeMathematics Jan 09 '23 at 13:51
  • @ILikeMathematics: I will reiterate mostly what KCd says. The limit in question is the derivative of $\log x$ at $x=1$ and you are welcome to find this derivative by any means (except L Hospital Rule because the rule assumes the derivative thing). – Paramanand Singh Jan 09 '23 at 15:31
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An elementary way to do the discrete version of this is to show that if $a_n=(1+1/n)^n$ and $b_n=(1+1/n)^{n+1}$ then $a_n$ is increasing and $b_n$ is decreasing and, as $n \to \infty$, $a_n$ and $b_n$ have a common limit, which I suggest we call $e$.

An elementary proof of this, which only uses the arithmetic mean-geometric mean inequality, can be found in N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563.

marty cohen
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