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Let {$x_n$} be monotone increasing sequence of positive real numbers. Show that if {$x_n$} is unbounded, then $\sum_{n=1}^{\infty}(1-\frac{x_n}{x_{n+1}})$ diverges.

Martin Sleziak
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user10024395
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Using the well-known inequality $\ln a\leq a-1$ with $a=x_{k+1}/x_k$ we obtain $$ \ln x_{k+1}-\ln x_k\leq \frac{x_{k+1}-x_k}{x_k} $$ Adding these inequalities for $k\in\{1,2,\dots,n-1\}$, we obtain $$ \ln x_{n}-\ln x_1\leq\sum_{k=1}^{n-1} \frac{x_{k+1}-x_k}{x_k} $$ but the sequence $\{x_n\}$ in increasing and unbounded, hence $$ \sum_{k=1}^{\infty} \frac{x_{k+1}-x_k}{x_k}=+\infty.\tag{1} $$ Now suppose, for contradiction, that the considered series does converge, $i.e.$ $$\sum\limits_{k=1}^{\infty} \dfrac{x_{k+1}-x_k}{x_{k+1}}<+\infty\tag{2}$$ then its general term must tend to zero, that is $\lim\limits_{k\to\infty}\dfrac{x_k}{x_{n+1}}=1$, which implies that $$ \dfrac{x_{k+1}-x_k}{x_{k+1}}\sim_{\infty}\dfrac{x_{k+1}-x_k}{x_{k}} $$ and contradicts (1). This contradiction proves that (2) is absurd and the series $\sum\limits_{k=1}^{\infty} \dfrac{x_{k+1}-x_k}{x_{k+1}}$ must be divergent.

Omran Kouba
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  • Same thing here. I dun understand why we can use approximation to compute limit – user10024395 Apr 08 '14 at 13:35
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    @user136266 your are not computing anything, you are proving the series diverges and approximation is a legit way of doing it. but if you don't like it you can try this approach http://math.stackexchange.com/questions/746257/prove-sum-infty-n-1-fraca-n-a-n-1a-n-infty – toufik_kh.17 Apr 10 '14 at 06:29
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If $\frac{x_n}{x_{n+1}}$ does not tend to $1$ then the sequence diverges because the $n$th term does not tend to zero. If $\frac{x_n}{x_{n+1}}\to 1$ then your sequence is equivalent to the sequence $\log(\frac{x_n}{x_{n+1}})$ which diverges by a telescoping series argument. The reason the sequence is equivalent to $\log(\frac{x_n}{x_{n+1}})$ is because as $x\to 0$ one has $\log (1+x)\sim x$.

Mikhail Katz
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  • I don't get how is the sequence equivalent to $log(x_n/x_{n+1})$. Where does boundedness come into play here? – user10024395 Apr 08 '14 at 11:52
  • why we can use approximation when finding limit? I mean i haven't learnt such a thing yet? Is it rigorous to do that? – user10024395 Apr 08 '14 at 13:35
  • Yes, if we have to positive sequences which are equivalent (more generally, differ by at most a certain fixed constant), then they will converge or diverge simultaneously. This is obvious because the constants provide a way of bounding each sequence in terms of the other. – Mikhail Katz Apr 08 '14 at 15:40
  • doesn't that mean that we have to find the fixed constant first to make the proof rigorous? – user10024395 Apr 08 '14 at 17:19
  • If $x\to 0$ then $\frac{\log (x+1)}{x}\to 1$ and therefore for $x$ sufficiently close to $0$ the ratio will be between 1/2 and 2. – Mikhail Katz Apr 08 '14 at 18:12