Let $\{x_n\}$ be monotone increasing sequence of positive real numbers. Show that if $\{x_n\}$ is bounded, then $\sum_{n=1}^{\infty}\left(1-\frac{x_n}{x_{n+1}}\right)$ converges. On the other hand, if the sequence is unbounded, the series is divergent.

The following is the proof given by my lecturer, but I do not understand the proof:

Let $u_m = x_{m+1} -x_m$ and $d_m = \sum_{k=1}^m u_k = x_{m+1} - x_1$ and hence $x_{m+1} = d_m + x_1$

$$\sum_{n=1}^{\infty}\left(1-\frac{x_n}{x_{n+1}}\right) = \sum_{n=1}^{\infty}\frac{u_m}{d_m +x_1} $$

Then he says that if $d_m$ diverges, then the sum diverges. If $d_m$ converges, then then sum converges.

He says that it has something to do with Gauss Test.

Can someone explain the proof to me. Or does anyone has a better proof.

  • $\begingroup$ I don't know if this is true. If $x_n \to 1$ the $n^{th}$ term test says $1-\frac{.5}{1.5}$ must be $0$, but it is not. $\endgroup$ – Cbjork Apr 2 '14 at 18:17
  • $\begingroup$ sorry it should be x_{n+1} $\endgroup$ – user10024395 Apr 2 '14 at 18:29

Since $1+a+b\le(1+a)(1+b)$, induction yields $$ 1+\sum_{n=1}^ma_n\le\prod_{n=1}^m(1+a_n) $$ Therefore, suppose $x_n$ is bounded. Then $$ \begin{align} 1+\sum_{n=1}^\infty\left(1-\frac{x_n}{x_{n+1}}\right) &\le1+\sum_{n=1}^\infty\left(\frac{x_{n+1}}{x_n}-1\right)\\ &\le\prod_{n=1}^\infty\frac{x_{n+1}}{x_n}\\[6pt] &=\lim_{n\to\infty}\frac{x_n}{x_1}\\ \end{align} $$ So, if $x_n$ is bounded, $\sum\limits_{n=1}^\infty\left(1-\frac{x_n}{x_{n+1}}\right)$ converges.

Suppose $\sum\limits_{n=1}^\infty\left(1-\frac{x_n}{x_{n+1}}\right)$ converges. Then $\frac{x_n}{x_{n+1}}\lt\frac12$ only finitely many times. Thus, there is an $N$ so that $x_n\ge\frac12x_{n+1}$ for $n\ge N$. Furthermore, $x\ge\log(1+x)$. Therefore, $$ \begin{align} \sum_{n=1}^\infty\left(1-\frac{x_n}{x_{n+1}}\right) &\ge\sum_{n=N}^\infty\left(1-\frac{x_n}{x_{n+1}}\right)\\ &\ge\frac12\sum_{n=N}^\infty\left(\frac{x_{n+1}}{x_n}-1\right)\\ &\ge\frac12\sum_{n=N}^\infty\log\left(\frac{x_{n+1}}{x_n}\right)\\ &=\frac12\lim_{n\to\infty}\log\left(\frac{x_n}{x_N}\right) \end{align} $$ Thus, if $\sum\limits_{n=1}^\infty\left(1-\frac{x_n}{x_{n+1}}\right)$ converges, $x_n$ is bounded.


Since {$x_n$} is monotone increasing of positive real numbers, i.e. $x_n > x_1$, we have

$$\sum_{n=1}^{M}(1-\frac{x_n}{x_{n+1}}) = \sum_{n=1}^{M}(\frac{x_{n+1}-x_n}{x_{n+1}})\leq \frac{1}{x_1} \sum_{n=1}^{M}x_{n+1}-x_n = \frac{1}{x_1} (x_M -x_1)$$

If {$x_n$} is bounded above by say M, then we have

$$\sum_{n=1}^{M}(1-\frac{x_n}{x_{n+1}}) \leq \frac{1}{x_1}(M-x_1)$$

Since $(1-\frac{x_n}{x_{n+1}})$ is non-negative and the sequence of partial sums $\sum_{n=1}^{M}(1-\frac{x_n}{x_{n+1}})$ is bounded, the sequence $\sum_{n=1}^{\infty}(1-\frac{x_n}{x_{n+1}})$ is convergent.

If {$x_n$} is unbounded, {$x_n$} is divergent. Using the inequality $\ln n \leq n$ (which can be proven by induction) and let $n = \frac{x_{n+1}}{x_n}$, we have

$$\ln \frac{x_{n+1}}{x_n} \leq \frac{x_{n+1}}{x_n}$$

Since $\sum_{n=1}^{M-1} \ln \frac{x_{n+1}}{x_n} = \sum_{n=1}^{M-1} \ln x_{n+1} - \ln x_n = \ln x_M - \ln x_1$. Since {$x_n$} is divergent, {$\ln x_n$} is divergent as well. This implies that $\sum_{n=1}^{M-1} \ln \frac{x_{n+1}}{x_n}$ is divergent. By comparison test, $\frac{x_{n+1}}{x_n}$ is divergent.

Since $\frac{x_{n+1}}{x_n}$ is divergent, for sufficiently large $n_0$, we will have $\frac{x_{n+1}}{x_n} > 2$. Therefore, we have

$$\sum_{n=n_0}^{\infty}(1-\frac{x_n}{x_{n+1}}) = \sum_{n=n_0}^{\infty}(1-\frac{1}{\frac{x_{n+1}}{x_n}}) > \sum_{n=n_0}^{\infty}(1-\frac{1}{2}) = \sum_{n=n_0}^{\infty}\frac{1}{2} \text{which diverges}.$$

Therefore, by comparison test, $\sum_{n=n_0}^{\infty}(1-\frac{x_n}{x_{n+1}})$ diverges.

  • $\begingroup$ In your answer for {x_n} unbounded, you suggest "By comparison test, x_{n+1}/x_n is divergent." How have you used the comparison test here? The sequence (x_n) given by x_n = n satisfies all the conditions, but x_{n+1}/x_n converges to 1 as n tends to infinity. $\endgroup$ – Zephos Mar 7 '15 at 14:52

I try to rewrite the proof cleaning some details.

Call $r_n=\frac{x_n}{x_{n+1}}\in(0,1]$. If there's an infinite number of terms such that $1-r_n=1-\frac{x_n}{x_{n+1}}\geq \frac12$ then the series clearly diverges, while by rewriting the condition as $x_{n+1}\geq 2 x_n$ also the sequence diverges.

If instead eventually $\frac12\leq r_n\leq 1$, then we can use that $$\log r\leq r-1\leq C\log r\qquad\text{for}\quad \frac12 \leq r\leq 1$$ for some positive $C$ to obtain $$\sum_{n=1}^N\left(\frac{x_n}{x_{n+1}}-1\right)\leq C \sum_{n=1}^N \log \frac{x_n}{x_{n+1}}=C\log\frac{x_1}{x_N}$$ and $$\sum_{n=1}^N\left(\frac{x_n}{x_{n+1}}-1\right)\geq \sum_{n=1}^N \log \frac{x_n}{x_{n+1}}=\log\frac{x_1}{x_N}.$$ Therefore the series has the same behaviour of $\lim_{N\to\infty}\log\frac{x_1}{x_{N}}$, which diverges iff $x_N\to\infty$.


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