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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?

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    $\begingroup$ No, you have $$\sum_{n = 1}^\infty (n-1)a_n > 0.$$ $\endgroup$ – Daniel Fischer Aug 7 '15 at 11:26
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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 $a_n=2^{-n}$.

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  • $\begingroup$ Thanks for the fast answer. The proof turned out to be rather easy indeed. :) $\endgroup$ – Keba Aug 7 '15 at 11:37
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    $\begingroup$ Just a minor correction: $$0 \le \sum_{n \ge 1} (n-1)a_n = \sum_{n \ge 1} na_n - \sum_{n \ge 1} a_n =0$$ implies that all terms of the sequence, EXCEPT $a_1$, are $0$. Any way it is a contradiction. $\endgroup$ – Ramiro Aug 7 '15 at 12:04
  • $\begingroup$ This assumes that the series is a positive series? $\endgroup$ – Belgi Aug 7 '15 at 12:45
  • $\begingroup$ @Belgi: Yes, $a_n \gt 0$ is a requirement in the question $\endgroup$ – Keba Aug 7 '15 at 18:07
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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$

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