5
$\begingroup$

Consider the function $$ f: (-1, 1) \rightarrow \mathbb{R}, \hspace{15px} f(x) = \sum_{n=1}^{\infty}(-1)^{n+1} \cdot \frac{x^n}{n} $$

I am required to show that the derivative of this function is $$ f'(x) = \frac{1}{1+x} $$

I have attempted to do this using the elementary definition of a limit, as follows: $$ f'(x) = \lim_{a \rightarrow x} \left( \frac{f(a)-f(x)}{a-x} \right) = \lim_{a \rightarrow x} \left( \frac{ \sum_{n=1}^{\infty} \left[ (-1)^{n+1} \cdot \frac{a^n}{n} \right]- \sum_{n=1}^{\infty}\left[ (-1)^{n+1} \cdot \frac{x^n}{n} \right]}{a-x} \right) \\ = \lim_{a \rightarrow x} \left( \frac{ \sum_{n=1}^{\infty} \left[ (-1)^{n+1} \cdot \frac{a^n - x^n}{n} \right]}{a-x} \right) $$

but I am unsure of how to proceed from here (assuming my approach is correct. Can someone help me to show this?

$\endgroup$
1
$\begingroup$

HINT: if you knows that

$$\int_0^x\frac{\mathrm dt}{1+t}=\ln(1+x),\quad x\in (-1,1)$$

then the question is equivalent to prove that

$$\sum_{k=1}^\infty(-1)^{k+1}\frac{x^k}k=\ln(1+x),\quad x\in(-1,1)$$

$\endgroup$
  • $\begingroup$ Someone can explain the downvote? $\endgroup$ – Masacroso Mar 7 '17 at 21:53
  • $\begingroup$ No, but I counter-voted, :) $\endgroup$ – Andres Mejia Mar 8 '17 at 9:43
0
$\begingroup$

Use the fact that $a^n-x^n=(a-x)\sum_{i=1}^na^{n-i}x^{i-1}$, canceling with the denominator and taking the limit as $a\to x$ we get $(-1)^{n+1}nx^{n-1}$ in the numerator. Cancel the $n$ and now prove that the resulting sum converges to $\frac{1}{1+x}$ using alternating series.

$\endgroup$
0
$\begingroup$

You could observe that this is the taylor expansion for $\ln(x+1)$ and take its first derivative, if you already know about the derivatives for $\ln(x+1)$.

Otherwise, you could note that the sequence of functions converges uniformly in its radius of convergence, and the derivative is a continuous and linear function, so

$\begin{align*}\frac{d}{dx} \lim_{k \to \infty} \sum_{n=1}^{k} (-1)^{n+1} \frac{x^n}{n}&=\lim_{k \to \infty} \sum_{n=1}^{k} (-1)^{n+1} \frac{d}{dx}\frac{x^{n}}{n}\\&=\lim_{k \to \infty}\sum_{n=1}^{k}(-1)^{n-1} n\cdot \frac{x^{n-1}}{n}\\&=\sum_{n=0}^{\infty}(-x)^n \end{align*}$

where the penultimate equality comes from the fact that $(-1)^{n+1}=(-1)^{n-1}$. This has the well known geometric series formula in the interval $(-1,1)$, and is precisely $\frac{1}{1+x}$.

$\endgroup$
0
$\begingroup$

No need to argue straight from the definition of the derivative. If you could show uniform convergence, you could simply differentiate term by term, which gives a nice geometric series.

So how do you show uniform convergence? Use the Weierstrass M test. On any closed interval $[-a,a]\subset(-1,1)$, the absolute values of the terms of your series are bounded by $a^n$, and because $|a|<1$, the series $\sum a^n$ converges. So your series converges uniformly on $[-a,a]$. This is all you need to get the derivative at any point $x_0\in(-1,1)$, for you can simply pick $a$ so that $|x_0|<a<1$.

$\endgroup$
  • $\begingroup$ (Or just note the radius of convergence of your power series is $1$, and invoke the result that a power series converges uniformly on any compact subset of its interval of convergence. But the proof of that result just uses the M test in exactly the same way I've used it above.) $\endgroup$ – symplectomorphic Mar 7 '17 at 22:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.