Existence of integer sequence for any positive real number I am stuck on the following problem:
Given any $\lambda > 0$ there exists strictly increasing sequence of positive integers  $$ 1 \leq  n_1 < n_2 < ...  $$ such that 
$$\sum _{i=1}^{\infty} \frac{1}{n_i} = \lambda$$
Any help will be appreciated.
Thank you.

You can use the following lemma : for any two sequence $u_{n}$ and $v_{n}$ from 
$\mathbb{R}^{\mathbb{N}}$ that tend to $+\infty$ , and $u_{n+1}-u_{n} \to 0$
then the set $\{u_{n}-v_{m}|(n,m)\in \mathbb{N}^{2}\}$ is dense in $\mathbb{R}$ . 
Proof : 
let $e>0$ there is $k$ such that $\forall n \geq k$ we have $|u_{n+1}-u_{n}| < e$ ,  now let $s\geq u_{k}$ the set 
$H=\{ n\geq k | s\geq u_{n}\}$ is a noempty ($k$ is in it) finite part of $\mathbb{N}$ since $u_{n} \to +\infty$ let 
$m=max(H)$  hence $u_{m}\leq s < u_{m+1}$ 
now let $x \in \mathbb{R}$ ,  since $v_{n} \to +\infty$ there is $p$ such that $x+v_{p} \geq u_{k}$
and from the above we know that there is $n$ such that $|u_{n}-(x+v_{p})|< e$  (just take $s=x+v_{p}$ ) 
which is $|u_{n}-v_{p}-x| < e$ Done !
thake $u_{n}=v_{n}=H_{n}$ 
