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Let $ {a_k}$ be an unbounded,strictly increasing sequence of positive real numbers and $x_k=\dfrac{a_{k+1}-a_k}{a_{k+1}}$. Prove that $\sum_{k=m}^n x_k>1-\dfrac{a_m}{a_n}\forall n\geq m $ and $\sum_{k=1}^\infty x_k$ diverges to $+\infty.$

What I thought: $\sum_{k=m}^n x_k=\sum_{k=m}^n(1-\dfrac{a_k}{a_{k+1}})$

As $0<\dfrac{a_k}{a_{k+1}}<1$,taking $\dfrac{a_k}{a_{k+1}}=1-\dfrac{1}{k}$ for each $m\leq k \leq n$.

But i cant carry it more .Please help me.

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$$\sum_{k=m}^n x_k=\sum_{k=m}^n {a_{k+1}-a_k \over a_{k+1}}>\sum_{k=m}^n {a_{k+1}-a_k \over a_{n+1}}={a_{n+1}-a_m\over a_{n+1}}=1-{a_m \over a_{n+1}}>1-{a_m \over a_{n}}$$

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  • $\begingroup$ It does not make a great difference, but the definition of $x_k$ has $a_{k+1}$ in the denominator, not $a_k$. $\endgroup$ – Martin R Dec 3 '15 at 15:49
  • $\begingroup$ Ah, thanks. I'll fix that. $\endgroup$ – Kay K. Dec 3 '15 at 15:49
  • $\begingroup$ Note that the question has actually been answered before (and the answer math.stackexchange.com/a/1544037/42969 to the "duplicate" seems to have the same minor error). $\endgroup$ – Martin R Dec 3 '15 at 15:52
  • $\begingroup$ Yeah I also saw the comment after I answered and could see somebody had answered it. It was a mere coincidence that I made a similar mistake though.. $\endgroup$ – Kay K. Dec 3 '15 at 15:55

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