$\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty n a_n$ possible? Recently I found (somewhere on math.se) a nice proof for $\sum_{n=0}^\infty \frac{n}{2^n} = 2$ and thought “oh, that‘s surprising, as also $\sum_{n=0}^\infty \frac{1}{2^n} = 2$ and it ‘feels like’ the first series should be greater than the later one.” However the surprise did not last very long, as I noticed the first summand of the first series is zero. So basically one has $\sum_{n=1}^\infty \frac{n}{2^n} = 2 > 1 = \sum_{n=1}^\infty \frac{1}{2^n}$, which is less surprising.
So I asked myself whether there exists a series $\sum_{n=1}^\infty a_n$ with 


*

*$\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty n a_n$

*$a_n \gt 0$ for $n \in \mathbb{N}$ (ie. no $0 = 0$ tricks)

*both series converge (ie. no $\infty = \infty$ tricks)


I tried $a_n = \frac{1}{n!}$, but then $\sum_{n=1}^\infty a_n = e-1 \lt e = \sum_{n=1}^\infty n a_n$ (interestingly, the difference is $1$ again!) and then I failed to find an example (or a proof that this is impossible).
Does such a series exist?
 A: Suppose such a series exists. Then
$$0 \le \sum_{n \ge 1} (n-1)a_n = \sum_{n \ge 1} na_n - \sum_{n \ge 1} a_n =0$$
so all terms of the sequence are $0$. A contradiction.
The problem is that you cannot make confusion with $\sum_{n \ge 1} na_n$ and $\sum_{n \ge 0} na_n$ because the second one has the first term equal to $0$, while the second no.
In general, under the hypothesis that $0<a_n$ and both sums are convergent,
$$\sum_{n \ge 0} na_n  = \sum_{n \ge 0} a_n$$
can be rewritten as 
$$a_1 + 2a_2 + 3a_3 + \dots = a_0 + a_1 + a_2 + a_3 + \dots$$
which is equivalent to
$$a_2 + 2a_3 + \dots = a_0$$
which is the case when you have $a_n=2^{-n}$.
A: Assum that $f(x)= \sum_{n \geq 0} a_n x^n$ and $f'(x)-f(x)-f(0)=0$ is true for $x\in[0,R]$ that $R$ is a Radius of convergence of $f(x)$. it is clearly that $R\geq1$ because $f(1)= \sum_{n \geq 0} a_n$.
solution of this ODE is our series.
$f(x)= \sum_{n \geq 0} a_n x^n, \quad f(1)= \sum_{n \geq 0} a_n , \quad f(0)=a_0,$
$f'(x)= \sum_{n \geq 1} na_n x^{n-1}, \quad f'(1)= \sum_{n \geq 1} na_n ,$
$\sum_{n \geq 1} na_n-\sum_{n \geq 0} a_n-a_0=f'(1)-f(1)-f(0)=0,$
Assume that $x\in[0,R]$:
$ \sum_{n \geq 1} na_n x^{n-1}-\sum_{n \geq 0} a_n x^n-f(0)=f'(x)-f(x)-f(0)=0$
