Does $\sum_{n=1}^{\infty} x_n$ converge when $x_{n+1}=\frac{e^{x_n}-1}{e^{x_n}}$ and $x_1 > 0$?

Question

Given $x_1>0$, and $$x_{n+1}=\frac{e^{x_n}-1}{e^{x_n}},$$determine whether the series $\sum_{n=1}^\infty x_n$ diverges. If it does, prove it; if not, provide a counterexample.

This is a series in iterative form, and it does not have a specific explicit form. The ratio test does not provide a conclusive result, and the root test does not seem suitable for this series. I feel a bit unfamiliar with this type of problem, and I would appreciate it if someone could provide some insights. Thank you very much.

Answer 1

$f(x)=1-e^{-x},g(x)=1-\frac{1}{x+1}$. It's obvious that $\forall x>0,f(x)>g(x)$.

Let $y_1=x_1,y_n=g(y_{n-1})$. We claim that $x_n\geq y_n$.

For the series $\{y_n\}$, we have $y_n=\frac{y_{n-i}}{iy_{n-i}+1}=\frac{y_1}{(n-1)y_1+1}=\frac{1}{n+c}$.

Since $\sum_{n=1}^\infty y_n=\sum_{n=1}^\infty \frac{1}{n+c}$ doesn't converge, $\sum_{n=1}^\infty x_n$ doesn't converge.

Answer 2

This method is a little long but can be used for a significantly large number of problems relating to recurrence relations. It consists of three steps :

• Prove that the required term goes to $0$ using an appropriate squeeze function.

• Find the rate at which the term goes to $0$ using the "physicist" method combined with Stolz-Cesaro and appropriate Taylor expansions.

• Do whatever calculations you need to with the rate(convergence of sum etc.)

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Convergence

First, prove that $x_n \to 0$. For this, note that $$ f(x)=\frac{e^x-1}{xe^x} ,x\in (0,1) \implies f'(x)<0\\ \lim_{x \to 0} f(x)=1 $$ Therefore, $f(x)$ is decreasing on $(0,1]$, hence $f(x) \leq \lim_{x \to 0}f(x) = 1$ for all $x\in (0,1)$. In particular, $$ \frac{e^x-1}{e^x} \leq x, x \in (0,1) $$

However, regardless of the value of $x_1>0$, we see that $x_i \in (0,1), i \geq 2$. As a consequence of the above inequality, $$ \frac{e^{x_n}-1}{e^{x_n}} \leq x_n, n \geq 2 \implies x_{n+1}\leq x_n , n \geq 2 $$

Thus, $x_n$ is a monotonic bounded sequence. It must converge to a point $L$ such that $$ \frac{e^{L}-1}{Le^L} = 1 \implies L=0 $$ (since $L\in[0,1]$ must occur, but we already saw above that $f(x) \neq 1$ for $x \in (0,1]$).

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Rate calculation : physicist method

We will now attempt to look at how fast $x_n$ converges to $0$. In particular, we will locate a function $n$ such that $f(n)x_n$ is at least bounded between positive constants as $n \to \infty$.

The "physicist" technique says the following : try to figure out $x_n$ as a function of $n$ by attempting to create a difference equation for $x_n$ which can be rephrased as a differential equation.

In this case, we know that $x_{n+1} = \frac{e^{x_n}-1}{e^{x_n}}$, so that $x_{n+1}-x_n = \frac{e^{x_n}-1}{x_n}-x_n$. The function $f(x)=\frac{e^x-1}{x}-x$ has the following property by Taylor expansion : $$ \lim_{x \to 0}\frac{\frac{e^x-1}{x} - x}{x^2} = -\frac 12 $$ In other words , $\frac{e^{x_n}-1}{x_n}-x_n \approx -\frac 12 x_n^2$ for large enough $n$. Therefore, $x_{n+1}-x_n \approx -\frac 12x_n^2$ for large enough $n$. As a consequence, if $x_n= f(n)$ where $f$ is some differentiable function, then the idea is that $x_{n+1}-x_n = f(n+1)-f(n) \approx f'(\xi)$ for some $\xi\in [n,n+1]$ by the mean value theorem, but assuming that $f'$ doesn't vary much, we simply write $f(n+1)-f(n) \approx f'(n)$. This leads to a simple differential equation $$ f'(n) \approx -\frac{1}{2}f(n)^2 \implies f(n) \approx \frac {1}{2n} $$

Therefore, we expect that $\frac{x_n}{\frac 1n}$ will be something bounded as $n \to \infty$. This is reflected in the following much stronger result.

$\lim_{n \to \infty} nx_n = 2$.

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Stolz-Cesaro

To prove the preceding result, we take advantage of the Stolz-Cesaro theorem. In particular, it says that $$ \lim_{n \to \infty} \frac{1}{nx_n} = \lim_{n \to \infty} \frac 1{x_n} - \frac{1}{x_{n-1}} $$

To calculate the former, we simple consider the function $$ g(x) = \frac{1}{\frac{e^{x}-1}{e^x}} - \frac 1x, x\in (0,1) $$ and ask for the limit $\lim_{x \to 0} g(x)$. Note that $g(x) = \frac{xe^x-e^x+1}{x(e^x-1)}$. A couple of L'Hospital applications reveal that $\lim_{x \to 0} g(x) =\frac 12$. In particular, $\lim_{ n \to\infty} g(x_n) = \frac 12$, and by choice of $g$, this means that $\lim_{n \to \infty} \frac{1}{x_n} - \frac{1}{x_{n-1}} = \frac 12$.

As a consequence, $\lim_{n \to \infty} nx_n = \frac {1}{\frac 12} = 2$.

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Final

It is now trivial to see that $\sum x_n$ doesn't converge. Indeed, by the definition of limits, there exists $N$ such that for $n>N$, $x_n > \frac{1}{3N}$. We are done by the comparison test for series.

Remark : Note that the above shows, for example, that $\sum_{n=1}^{\infty} x_n^2$ converges. Furthermore, one can use versions of Stolz-Cesaro and Taylor expansions of higher orders to study more intricate behaviors related to $x_n$.

Note : Also see here for a much harder problem solved in the exact same way, but requiring more work.

Answer 3

Using $e^u \ge 1+ u$, you can easily prove that the the sequence is non-increasing. It is not hard (using induction) to prove that the sequence is non-negative. This proves that the sequence converges. Let $\ell$ its limit so $$\ell = 1-e^{-\ell}$$ this proves that $\ell$ satisfies $g(\ell) =0$ with $g(x) = e^{-x} - 1 + x$. By studying the variation of $g$ you can easily show that $\ell =0$. Now with this you will have:

$$x_{n+1} = x_n - \frac12 x_n^2 + o\left(x_n^2\right)$$

Then, $$\frac1{x_{n+1}} = \frac1{x_n} + \frac12 + o(1)$$

Use Cesaro Lemma to prove that

\begin{align} \lim\limits_{n\to\infty} \frac1{nx_n} &= \lim_{n\to\infty} \frac1{x_{n+1}} - \frac1{x_n} = \frac12 \end{align}

Can you conclude from here?

Answer 4

This is essentially entoryvekum's solution, just written a bit differently:

All $x_n$ are positive, and we have $$ \frac{1}{x_{n+1}} = \frac{e^{x_n}}{e^{x_n}-1} = 1 + \frac{1}{e^{x_n}-1} \le 1 + \frac{1}{x_n} $$ using the well-known estimate $e^u \ge 1+u$. It follows that $$ \frac{1}{x_n} \le (n - 1) + \frac{1}{x_1} \implies x_n \ge \frac{x_1}{(n-1)x_1 + 1} $$ and therefore the divergence of $\sum x_n$.