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Consider the alternating series $$\phi(x):=\sum_{n=1}^{\infty}\dfrac{(-1)^{n}}{\sqrt{n}(1+\frac{x^{2}}{n})^{n}}.$$ The exercise asks me to show that this series does not converge absolutely on $\mathbb{R}$, converges uniform on $\mathbb{R}$ and $\phi(x)$ is differentiable on $\mathbb{R}$.

I have proved the first two, but got stuck in the last one. To show $\phi(x)$ is differentiable on $\mathbb{R}$, we need to show that $$(1)\ \ u_{n}(x):=\dfrac{(-1)^{n}}{\sqrt{n}(1+\frac{u^{2}}{n})^{n}}\ \ \text{is differentiable on}\ \ [-R,R];$$ $$(2)\ \ \sum_{n=1}^{\infty}u_{n}'\ \ \text{converges uniformly to some}\ \ g\ \ \text{on}\ \ [-R,R];$$ $$(3)\ \ \sum_{n=1}^{\infty}u_{n}(x_{0})\longrightarrow\phi(x_{0})\ \ \text{for some}\ \ x_{0}\in[ -R,R],$$ where $R$ is arbitrarily fixed. Then, $\phi$ is differentiable on $[-R,R]$ and $\phi'=g$. But $R$ is arbitrary, so taking $R\rightarrow\infty$ finishes the proof.

$(1)$ is clear and the derivative is $$u_{n}'(x)=\dfrac{(-1)^{n+1}2x}{\sqrt{n}(1+\frac{x^{2}}{n})^{n+1}},\ \ \text{for}\ \ n\geq 1.$$

$(3)$ is also clear. Since $(1+\frac{x^{2}}{n})^{n}\nearrow e^{x^{2}}$ and $\sqrt{n}\nearrow \infty$, we have $\frac{1}{\sqrt{n}(1+\frac{x^{2}}{n})^{n}}\searrow 0.$ Hence the alternating series converges pointwise on the whole $\mathbb{R}$ by the alternating series test.

However, I don't know how to show $(2)$. I want to use the following fact:

If a alternating series passes the alternating series test, then we must have $$\Big|\sum_{k=n}^{\infty}a_{k}\Big|\leq |a_{n}|.$$

I used this fact to prove the uniform convergence. However, the denominator of $u_{n}'$ does not have the same monotonicity:

$\sqrt{n}\nearrow\infty$, but $(1+\frac{x^{2}}{n})^{n+1}\searrow e^{x^{2}}$. Thus, even though on $[-R,R]$, $$\frac{2x}{\sqrt{n}(1+\frac{x^{2}}{n})^{n+1}}\rightarrow 0,$$ it is unknown that if this convergence is monotonically decreasing, but this is required by the alternating series test.

Is there any other way to prove the uniform convergence of $\sum u_{n}'$? or I am missing something?

Thank you!

Edit 1:

Below is how I proved the uniform convergence of $\sum_{n=1}^{\infty}u_{n}(x)$.

Let $\epsilon>0$, take $N:=\frac{1}{\epsilon^{2}}-1$, then as the alternating series passes the alternating series test, we use the referred fact above, and see that for all $n\geq N$, and for all $x\in\mathbb{R}$, we have \begin{align*} \Big|\sum_{k=n+1}^{\infty}u_{k}(x)\Big|\leq |u_{n+1}(x)|&=\Big|\dfrac{(-1)^{n+1}}{\sqrt{n+1}(1+\frac{x^{2}}{n+1})^{n+1}}\Big|\\ &=\dfrac{1}{\sqrt{n+1}}\Big|\dfrac{1}{(1+\frac{x^{2}}{n+1})^{n+1}}\Big|\\ &\leq\dfrac{1}{\sqrt{n+1}}\\ &\leq \dfrac{1}{\sqrt{N+1}}\\ &=\dfrac{1}{\sqrt{\frac{1}{\epsilon^{2}}-1+1}}\\ &=\epsilon. \end{align*}

As I argued before, this proof requires the alternating series to pass the alternating series test, but the coefficient of $u_{n}'(x)$ may not decrease to $0$. (It indeed goes to $0$).

Arctic Char
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JacobsonRadical
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  • How did you show that $\sum_{n=1}^\infty u_n(x)$ converges uniformly on $[-R,R]$? If you have show that, then it should not be difficult to show that $\sum_{n=1}^\infty u_n'(x)$ converges uniformly on $[-R,R]$ also. One more thing … $\left(1+\frac{x^2}n\right)^{n+1}$ does not monotonically increase to $e^{x^2}$ for all $x$. – Mark Viola Jun 23 '20 at 21:11
  • @MarkViola I used the fact I mentioned in this post, and bound the tail series, using $N=\frac{1}{\epsilon^{2}}-1$. But to use this fact, I have to show $a_{n}(x)$ monotonically decreases to $0$. This works for $u_{n}(x)$, but as I argued in the post, this doest not work for $u_{n}'(x)$. – JacobsonRadical Jun 23 '20 at 21:15
  • @MarkViola Let me edit the post to add how I proved the uniform convergence. – JacobsonRadical Jun 23 '20 at 21:21
  • @MarkViola edited just now. – JacobsonRadical Jun 23 '20 at 21:26
  • Jacob, it is not correct that $\left(1+\frac{x^2}n\right)^{n+1}$ monotonically decreases to $e^{x^2}$. For example, take the case $x=2$. Then, $\left(1+\frac{4}n\right)^{n+1}\le e^4$ for all $n\ge 1$. But $\left(1+\frac{x^2}n\right)^{n+1}$ does not monotonically increase to $e^{x^2}$ for all $x$ either. Take $x=1$ for example. – Mark Viola Jun 24 '20 at 02:12
  • @MarkViola ah. okay. Thank you! I assume this due to the accepted answer here: https://math.stackexchange.com/questions/3720615/show-that-f-nx-nx1-xn-is-uniformly-bounded-on-0-1-for-all-n-geq – JacobsonRadical Jun 24 '20 at 14:46
  • You're welcome. The relationship is true for $x=1$, but not true in general. – Mark Viola Jun 24 '20 at 18:09
  • @MarkViola do you know if $(1+\frac{x}{n})^{n+1}$ converge to $e^{x}$? Perhaps it converges but just not monotonically? – JacobsonRadical Jun 24 '20 at 18:46
  • Yes Jason. See This Answer I posted in 2015. – Mark Viola Jun 24 '20 at 18:50
  • Jason, the sequence $\left(1+\frac xn\right)^n$ increases monotonically for all $x>-n$ and converges to $e^x$. But interestingly, we cannot assert that the sequence $\left(1+\frac xn\right)^{n+1}$ is monotonic for all $n$ for every fixed $x$. There are values of $x$ for which it increases, values of $x$ for which it decreases, and values of $x$ for which is not monotonic for all $n$. Try analyzing these two distinct sequences and supplement by plotting them. Let me know what you find. ;-) – Mark Viola Jun 24 '20 at 18:59
  • @MarkViola okay. I don't really know this concept you such a deep way. I will analyze them and come back to you. – JacobsonRadical Jun 24 '20 at 19:07
  • @MarkViola Thank you so much !! – JacobsonRadical Jun 24 '20 at 19:07
  • You're welcome Jason. My pleasure. And have a look at the answer in the link I provided to you. Here is another one. – Mark Viola Jun 24 '20 at 19:08
  • @MarkViola no problem. Thank you! – JacobsonRadical Jun 24 '20 at 19:11

2 Answers2

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Note that $2x/\sqrt{n}$ is monotonically decreasing with respect to $n$ and uniformly convergent to $0$ on $[0,R]$, We also have $-2x/\sqrt{n}$ montonically decreasing with respect to $n$ and uniformly convergent to $0$ on $[-R,0)$. Hence, the series

$$\sum_{n=1}^\infty \frac{(-1)^{n+1}2x}{\sqrt{n}}$$

converges uniformly by Dirichlet's test on both $[-R,0)$ and $[0,R]$, and, therefore, on $[-R,R]$.

The sequence $(1 + x^2/n)^{-(n+1)}$ is eventually monotone and uniformly bounded for $x \in [-R,R]$. Therefore, by Abel's test we have uniform convergence of

$$\sum_{n=1}^\infty \frac{(-1)^{n+1}2x}{\sqrt{n}} \cdot \frac{1}{\left(1 + \frac{x^2}{n} \right)^{n+1}}$$

RRL
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  • Thank you. Nice answer. – JacobsonRadical Jun 23 '20 at 21:29
  • @MarkViola Hope you also stay safe and healthy. Thanks for your help :) – JacobsonRadical Jun 23 '20 at 21:33
  • @rrl There is a problem here. Note that $a_n(x)=\left(1+\frac{x^2}n\right)^{n+1}$ does not monotonically increases for all $x$. For example, take $x=1$ and surely $a_{n+1}(1)\le a_n(1)$. Hence, its reciprocal does not satisfy the condition of Abel's test for all $x\in [-R,R]$ for all $R$ (that is, uniformly bounded AND decreasing). – Mark Viola Jun 23 '20 at 21:47
  • @MarkViola: It only needs to be monotone (increasing or decreasing). For ordinary convergence that only has to happen eventually. So it may be a snag for uniform convergence but I have to check. – RRL Jun 23 '20 at 22:12
  • I believe that Abel requires the second term (in the product that forms the general term of the series) to be monotonic and decreasing. This is not true here for all $x$. – Mark Viola Jun 23 '20 at 22:39
  • Here is a statement of Abel's test for UC of a series. Indeed, if $g_n(x)$ is monotonically decreasing for $x\in E$ and if $\sum_n f_n(x)$ UC on $E$, then $\sum_n g_n(x)f_n(x)$ is UC on $E$. – Mark Viola Jun 23 '20 at 22:49
  • I know that is sufficient. But there is another condition when $\sum |g_n(x) - g_{n+1}(x)|$ is uniformly convergent. This the case when the sequence $g_n$ is uniformly convergent and monotone so that $|g_n - g_{n+1}| = (g_n - g_{n+1}$ and the sum is telescoping. – RRL Jun 23 '20 at 23:00
  • Using summation by parts, $\sum_{k=1}^n f_k(x) g_k(x) = S_n(x) g_{n+1}(x) + \sum_{k=1}^nS_n(x)(g_n(x) - g_{n+1}(x))$. Here $S_n(x) = \sum_{k=1}^n f_k(x)$ converges uniformly to $S$ and $g_n(x)$ converges uniformly. So UC is guaranteed if $\sum_{k=1}^nS_n(x)(g_n(x) - g_{n+1}(x))$ converges uniformly. Since $S_n(x)$ is uniformly bounded this reduces to showing that $\sum_{k=1}^n|g_n(x) - g_{n+1}(x)|$ is uniformly convergent. – RRL Jun 23 '20 at 23:05
  • @RRL Alas, we don't have monotonicity of the $g_n$ sequence (not for all $x$). But you assert that $g_n$ will be monotone eventually, correct? That should be trivial to show. – Mark Viola Jun 23 '20 at 23:17
  • We also have $x$ in a compact set. – RRL Jun 23 '20 at 23:35
2

Knowing that $\sum_{n=1}^{\infty} u_n(x)$ converges locally uniformly, here is a quick proof: Write

\begin{align*} u_n'(x) = -\frac{2x}{1+\frac{x^2}{n}} u_n(x) = -2x u_n(x) + \frac{2x^3}{n+x^2} u_n(x). \end{align*}

Then on each interval $[-R, R]$,

  • $\sum_{n=1}^{\infty} (-2x) u_n(x)$ converges uniformly, and

  • $\left| \frac{2x^3}{n+x^2} u_n(x) \right| \leq \frac{2R^3}{n^{3/2}} $ uniformly in $n$ and $x$, and so, $\sum_{n=1}^{\infty} \frac{2x^3}{n+x^2} u_n(x)$ converges uniformly by the Weierstrass M-test.

Therefore $\sum_{n=1}^{\infty} u_n'(x)$ also converges uniformly on $[-R, R]$.

Addendum. Another purpose of this answer is to hint a much more general idea: You may split the sum into two parts: conditionally convergent part (which is easier to control) and absolutely convergent part.

In OP's case, we easily check that, for each $R> 0$,

$$ \frac{1}{(1+\frac{x^2}{n})^{n+1}} = e^{-x^2} + \mathcal{O}\left(\frac{1}{n}\right) $$

uniformly in $n$ and $x \in [-R, R]$, where the implicit bound of $\mathcal{O}(\frac{1}{n})$ depends only on $R$. Using this, we may write

$$ u_n'(x) = \frac{(-1)^{n+1}2xe^{-x^2}}{\sqrt{n}} + \mathcal{O}\left(\frac{1}{n^{3/2}}\right), $$

which can be used to easily prove the uniform convergence of $\sum_{n=1}^{\infty} u_n'(x)$ over $[-R, R]$.

Sangchul Lee
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