the debut problem

Prob1:
03.04.2019

Prove that : there is $a_{n}>0$ such that :

for all positive integers $k_{1},...,k_{n} \geq 1$ such that $\frac{1}{k_{1}}+...+\frac{1}{k_{n}} < 1$ then :

$\frac{1}{k_{1}}+....+\frac{1}{k_{n}} < 1-a_{n}$

Clue: Use supremum property .


Solution(posted on 10.04.19) :
Here is the updated solution:
As, $\sum_{i=1}^{n} \frac{1}{k_{i}}<1$
The set {$\sum_{i=1}^{n} \frac{1}{k_{i}}$:$\sum_{i=1}^{n} \frac{1}{k_{i}}<1$}$_{k_{i}\in N }$has a supremum,say,$S_{n}(<1)$...(∗)
(Assuming)
Now,$S_{n}>= \sum_{i=1}^{n} \frac{1}{k_{i}}$
$\implies 0<t_{n}$(say)$=1-S_{n}<=1-\sum_{i=1}^{n} \frac{1}{k_{i}}$
Now,by Archimedean principle ,
$\exists N_{n}$ such that $\frac{1}{N_{n}}<t<1-\sum_{i=1}^{n} \frac{1}{k_{i}}$
Now,take,$a_{n}= \frac{1}{N_{n}} $ which is unique,giving us,
$\sum_{i=1}^{n} \frac{1}{k_{i}}<1-a_{n}$.
So proving (∗) will complete the proof.

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