Sequence {$x_n$} of positive terms monotonically increasing and bounded.

How to prove that $\sum_{n=1}^\infty \Big(1 - \frac{x_n}{x_{n+1}} \Big)$ is convergent?

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    $\begingroup$ First show that $x_n$ converges to some $x > 0$. Then try to show that we have $1 - \frac{x_n}{x_{n+1}} \leq C[x_{n+1} - x_n]$ for some constant $C > 0$. $\endgroup$ – Winther Nov 15 '16 at 2:04
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    $\begingroup$ Good idea! Some similar questions. $\endgroup$ – A.Γ. Nov 15 '16 at 2:20

First, since $x_n$ is positive, increasing, and bounded, the limit, $\lim_{n\to \infty}x_n$, exists. Denote this limit by $x_{\infty}$.

Now, let $s_n=1-\frac{x_n}{x_{n+1}}$. Clearly, since $x_n\le x_{n+1}$, then $s_n\ge 0$. Hence, the sequence of partial sums, $S_N=\sum_{n=1}^Ns_n$, is monotonically increasing.

Next, we note that

$$0\le s_n= 1-\frac{x_n}{x_{n+1}}=\frac{x_{n+1}-x_n}{x_{n+1}}\le \frac{x_{n+1}-x_n}{x_2}$$

Therefore, we have

$$0\le S_N=\sum_{n=1}^N \left(1-\frac{x_n}{x_{n+1}}\right)\le \frac{x_{N+1}-x_1}{x_2}\to \frac{x_{\infty}-x_1}{x_2}$$

which shows that $S_N$ is bounded above. Therefore, inasmuch as $S_N$ is increasing and bounded above, it converges.

And we are done!

  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$ – Mark Viola Nov 15 '16 at 21:09
  • $\begingroup$ $$ x_1 \lt x_{n+1} \Rightarrow \frac{1}{x_{n+1}} \lt \frac{1}{x_1} \Rightarrow \frac{x_{n+1}-x_n}{x_{n+1}} \lt \frac{x_{n+1}-x_n}{x_1} \Rightarrow \small \sum_{n=1}^{N}\frac{x_{n+1}-x_n}{x_{n+1}} \lt \frac{1}{x_1}\sum_{n=1}^{N}(x_{n+1}-x_n) \\ \sum_{n=1}^{N}(+x_{n+1}-x_n)=\small (-x_1+x_2)+(-x_2+x_3)+(-x_3+x_4)+...+(-x_{\tiny N-1\small}+x_{\tiny N\small}) \normalsize = x_{\small N\normalsize}-x_1 \Rightarrow \\ \sum_{n=1}^{\infty}\frac{x_{n+1}-x_n}{x_{n+1}} \lt \frac{1}{x_1} \lim_{N\rightarrow\infty}(x_{\small N\normalsize}-x_1)=\frac{1}{x_1} (x_{\infty}-x_1)=\frac{x_{\infty}}{x_1}-1 $$ $\endgroup$ – Hazem Orabi Nov 15 '16 at 21:47
  • $\begingroup$ Greetings Mark, No question at all. I am just trying to explain your (good) answer. Thanks. $\endgroup$ – Hazem Orabi Nov 15 '16 at 21:58
  • $\begingroup$ OK. And thank you! Much appreciative. Of course feel free to up vote as you see fit. $\endgroup$ – Mark Viola Nov 15 '16 at 22:15

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