# Properties of $x_k=\frac{a_{k+1}-a_{k}}{a_{k+1}}$ where $\{a_n\}$ is unbounded, strictly increasing sequence of positive reals

Let $\{a_n\}$ be an unbounded, strictly increasing sequence of positive real numbers and let $x_k=\frac{a_{k+1}-a_{k}}{a_{k+1}}$. Which of the following statements is/are correct? (CSIR NET December 2014)

1. For all $n\geq m, \sum\limits^{n}_{k=m}x_k>1-\bf{\frac{a_m}{a_n}}$
2. There exist $n\geq m, \sum\limits^{n}_{k=m}x_k>\frac{1}{2}$

My attempt:

Analysing question in light of $a_k=k\implies x_k=\frac{k+1-k}{k+1}=\frac1{k+1}$, I feel both are correct, I doubt if I can generalise the result in light of this one example.

Any way to prove that this true for general case?

Note: Please do not confuse this question with my question. The first two options are different, I've marked the difference in bold.

• Right, these are formally different, but the proofs are carbon-copies. What is stopping you? – Did Nov 24 '15 at 7:13
• @Did, quoting answer, First option is wrong there, but true here. – Jesse P Francis Nov 24 '15 at 7:17

As $\{a_n\}$ is strictly increasing and positive, $$\sum_{k=m}^n x_k = \sum_{k=m}^n \frac{a_{k+1}-a_k}{a_{k+1}} \ge \sum_{k=m}^n \frac{a_{k+1}-a_k}{a_{n+1}} = \frac{a_{n+1} - a_m}{a_{n+1}} \ge 1 -\frac{a_m}{a_{n+1}}\ge 1-\frac{a_m}{a_n}.$$
Since $a_n\to \infty$, for all $m$ there is $n>m$ so that $a_n > 2a_m$, so $$\sum_{k=m}^n x_k > \frac 12.$$
• I hope you meant summation from $k=m$ to $n$ in RHS of first inequality? – Jesse P Francis Nov 24 '15 at 7:20
• Minor nitpick: The definition of $x_k$ has $a_{k+1}$ in the denominator, not $a_k$. – Martin R Dec 3 '15 at 15:59