Concerning series of positive real numbers whose terms are decreasing and tending to $0$ Let $\{a_n\}$ be a decreasing sequence of positive real numbers such that $\lim_{n \to \infty} a_n=0$ and 
$\sum_{n=1}^{\infty}a_n= \infty$ ( for eaxmple , like $a_n:=\dfrac 1n$ ), then is it true that for every $r>0$ , there exists a 
subsequence  $\{a_{r_n} \}$ of $\{a_n\}$ such that $\sum_{n=1}^{\infty} a_{r_n}=r$ ? For example , for $\sum\dfrac 1n =\infty$ and  $e>0$ , we have subsequence $\{\dfrac 1{n!}\}$ of $\{\dfrac 1n\}$ such that $\sum \dfrac 1{n!}=e$ , but even for this sequence $\{\dfrac1 n \}$ , I am not able to prove my  claim for general $r>0$ . Please help . Thanks in advance 
 A: We will split the proof in several propositions:


*

*There exists a subsequence $a_{n_k}$ such that $\sum_{k\ge 1} a_{n_k}<r$. The proof here is quite obvious, because we put $$n_1 = \min\{m\ge 1: a_m\le r/2\}\\ n_{k+1}=\min\{m> n_k: a_m\le 2^{-k-1}r \}.$$
Given this choice of $n_k$ we obtain that $\sum_{k\ge 1} a_{n_k}\le \sum_{k\ge 1} 2^{-k-1}r=r$.
In other words, the set of subsequences whose sum is below $r$ is non-empty.

*Suppose that we have a subsequence $a_{n_k}$ of finite sum, then the set of unused indices $U\subset \Bbb N$ is infinite. Moreover, the sum over $U$ is infinite. The proof is also obvious: indeed, if the set of unused indices is finite, then $\sum_{j\in U} a_j<+\infty$, and $\sum_{j\ge 1} a_j = \sum_{k\ge 1}a_{n_k}+\sum_{j\in U} a_j <\infty$, which is contradiction. If the sum over $U$ is finite, then absolute convergence of series $\sum_{j\in U} a_j$ and $\sum_{k\ge 1}a_{n_k}$ allows to say that $\sum_{j\ge 1} a_j = \sum_{k\ge 1}a_{n_k}+\sum_{j\in U} a_j <\infty$, which is also a contradiction.

*Denote $$f(s,l)=\sum_{j=s}^{s+l}a_j.$$ We claim that for any $\delta>0$ there exist $s$ and $l$ such that $$\delta/2\le f(s,l)\le \delta.$$
Indeed, let $s=\min\{m\ge 1: a_m\le \delta/2\}$. Note that $\lim_{l\to\infty}f(s,l)=+\infty$, because the series diverges. Also note that $f(s,1)<\delta/2$. Therefore there exists $l\ge 1$ such that 
$$f(s,l)\le \delta< f(s,l+1). $$
By monotonicity of the sequence we get $a_{s+l+1}\le a_s\le \delta/2$, hence $$f(s,l)= f(s,l+1)-a_{s+l+1}\ge \delta-\delta/2=\delta/2, $$ which is what we needed to show.
Now we can put all these results to good use. We consider the set of all possible subsequences whose sum is inferior of equal to $r$. This set is non-empty by the first result. Now take any such subsequence $a_{n_k}$ and consider its sum $h$. If $h=r$, then we are fine. If $r-h=\delta>0$,  then we apply the third result to the set of unused indices $U$ and obtain  that we can include a finite number of terms in our subsequence such that the new subsequence $a_{n_k'}$ has that sum $h'$ in the bounds $[r-\delta/2,r]$.
We iterate this argument for $\delta_1=\delta/2$ to obtain a new subsequence with the sum in $[r-\delta/4,r]$, etc. In the end we will obtain the desired subsequence with the sum equal to $r$.
A: Let $r>0$.
1) Let $\epsilon_0=\min\{r,1\}$, and choose $N_1\in\mathbb{N}$ such that $a_n<\epsilon_0$ for $n\ge N_1$ .
Let $m_1$ be the smallest integer $m\ge0$ such that $\displaystyle s_1=\sum_{k=0}^{m}a_{N_{1}+k}>r-\epsilon_0$, so $r-1<s_1<r$.

(If $m_1=0$, then $s_1=a_{N_1}<\epsilon_0\le r;\;$ and 
if $m_1>0$, then $\displaystyle s_1=\sum_{k=0}^{m_1}a_{N_1+k}=\sum_{k=0}^{m_1-1}a_{N_1+k}+a_{N_1+m_1}<(r-\epsilon_0)+\epsilon_0=r$.)

2) Let $\epsilon_1=\min\{r-s_1,\frac{1}{2}\}$, and choose $N_2\in\mathbb{N}$ such that $N_2>N_1+m_1$ and $a_n<\epsilon_1$ for $n\ge N_2$.
Let $m_2$ be the smallest integer $m\ge0$ such that $\displaystyle s_2=s_1+\sum_{k=0}^{m}a_{N_{2}+k}>r-\epsilon_1$, so $r-\frac{1}{2}<s_2<r$.
3) Let $\epsilon_2=\min\{r-s_2,\frac{1}{3}\}$, and choose $N_3\in\mathbb{N}$ such that $N_3>N_2+m_2$ and $a_n<\epsilon_2$ for $n\ge N_3$.
Let $m_3$ be the smallest integer $m\ge0$ such that $\displaystyle s_3=s_2+\sum_{k=0}^{m}a_{N_{3}+k}>r-\epsilon_2$, so $r-\frac{1}{3}<s_3<r$.
Continuing in this manner by induction, we get a sequence $(s_n)$ such that $s_n\to r$, so  
the terms appearing in the $(s_n)$ give a subsequence $(a_{r_n})$ of $(a_n)$ with $\displaystyle \sum_{n=1}^{\infty}a_{r_n}=r$.

(Notice that this uses that $a_n>0$, $a_n\to 0$,  and that $\displaystyle\sum_{n=k}^{\infty} a_n$ diverges for any k, 
but not
that the sequence $(a_n)$ is decreasing.)
A: Let $$r= M + \sum_{i=1}^\infty \frac{b_i}{10^i},\ M,\ b_i\in \mathbb{Z},\ 0\leq M,\ 0\leq b_i
\leq 9
$$
(1) Find an smallest $N_0$ s.t. $$ a_{N_0}< 1 $$
And there exists $L_0\geq N_0$ s.t. $$ M-1\leq \sum_{i=N_0}^{L_0}
a_i < M$$ We contain these $a_i$ into a subsequence we desired.
(2) Find an smallest $N_1> L_0$ s.t. $$ a_{N_1}< \frac{1}{10}
$$
And there exists $L_1\geq N_1$ s.t. $$ M + \frac{b_1-1}{10} \leq
\sum_{i=N_1}^{L_1} a_i < M + \frac{b_1}{10} $$ We contain these
$a_i$ into a subsequence we desired.
We proceed these steps infinitely.
