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 

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.)
