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.
 A: Hint 1: if $\lambda\leq 1$, then you can use the binary representation of the real number  $\lambda $. (All your  $n_i $ will be powers of 2 in this case, therefore they are all even. This note is important for hint 2 to work)
Hint 2:if $\lambda >1$, then use the fact that the "odd" harmonic series diverges. i.e. $\sum_{i\in\mathbb{Z}^+, i \,\,\text{is odd}}=\infty$ to get a positive integer  $k $ such that:
$$0\leq \lambda -  (\sum_{i\in\{1,2,...,k\},i \,\,\text{is odd}}\frac{1}{i}) \leq 1$$
And then apply hint 1 to $\lambda -  (\sum_{i\in\{1,2,...,k\},i \,\,\text{is odd}}\frac{1}{i}) $
$$-------------------------------------------------------------$$
Thanks for Martin-Blas Perez Pinilla for spotting an error in my original proof.
A: For the case $\lambda > 1$ (the other case is covered in the answer of Amr) and excluding the trivial case $\lambda =$ finite sum of inverses: start taking $m_0$ s. t.
$$1+\frac12+\cdots+\frac1{m_0}=H_{m_0}<\lambda<H_{m_0+1}=1+\frac12+\cdots+\frac1{m_0+1}.$$
Obviously, $0<\lambda-H_{m_0}<1/(m_0+1)$. Now, take the minimum $m_1>m_0$ s.t.
$$0<\lambda-H_{m_0}<\frac2{m_1},\hbox{ i.e., }0<\lambda-H_{m_0}-\frac1{m_1}<\frac1{m_1}$$
(some exists because $0<\lambda-H_{m_0}<2/(2m_0+2)$)
Continue the process with
$$0<\lambda-H_{m_0}-\frac1{m_1}-\frac1{m_2}<\frac1{m_2}$$
...
Why the process is convergent?
A: 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 $u_{n}=v_{n}=H_{n}$ 
