If $\sum p_n = 1$, can $\sum np_n$ be unbounded? This question might seem quite easy and obvious, anyhow, I intend to make it clearer since this is just the beginning of a solution of a more general question.
Given $\sum_{i=0}^\infty p_i=1$ and $p_i>0$
What is the answer to this statement:
$1+p_1+2p_2+3p_3+4p_4+...=?$
Could it be infinity?
 A: For example $\sum \frac {1}{2^n} =1$ then $\sum \frac{n}{2^n}=2$.
But if we take the triangular numbers ${n+1\choose 2}$, we have that $\frac {1}{2} \sum \frac {1}{{n+1\choose 2}}=1$ but $\frac {1}{2}\sum \frac {n}{{n+1\choose 2}}=\sum \frac {1}{n+1}=\infty$
A: Actually $ p_i $ can be constructed. Let $a_k=\sum\nolimits_{i = k}^\infty  {{p_i}}$  , thus $a_0=1$, the sum turns out to be
$1+p_1+2p_2+3p_3+4p_4+\cdots = \sum\nolimits_{i = 0}^\infty  {{a_i}}$
If you want the sum to be finite, you should find a convergent sequence $(a_n) $. For example, let $a_i=\frac{1}{(i+1)^2}$, then
\begin{eqnarray}
a_n&=&\frac{1}{(n+1)^2}=\sum\nolimits_{i = n}^\infty  {{p_i}} \\
a_{n+1}&=&\frac{1}{(n+2)^2}=\sum\nolimits_{i = n+1}^\infty  {{p_i}} \\
p_n&=&a_n-a_{n+1}=\frac{1}{(n+1)^2}-\frac{1}{(n+2)^2}>0
\end{eqnarray}
That's the $p_i$ you've been looking for!
A: There is an easy way to see that $\sum_ip_i=1$ with $p_i\geq0$ tells you very little about $\sum_iip_i$. (I know the question says $p_i>0$, but that is just a nuisance to make counterexamples harder to formulate; any example with nonnegative terms can be transformed into one with positive terms by adding a sufficiently small amount to all zero terms; the effect this change has on $\sum_ip_i$ and $\sum_iip_i$ can be made as small as you like, and a slight rescaling will ensure that $\sum_ip_i=1$ remains true.)
Take any sequence with $\sum_{i\geq0}p_i=1$, say for instance $p_i=2^{-i-1}$. Now $\sum_iip_i$ might converge (it does in the example), but you can find something that increases more rapidly than $i$ to make it diverge; in the example for instance $\sum_i2^ip_i$ clearly diverges. Now to turn the converging example into a diverging one, simply spread out the sequence so that the term that used to be at place $i$ now is place at position $2^i$. So define a new sequence with $p'_{2^i}=2^{-i-1}$ and all other terms $p'_j$ (whose index$~j$ is not a power of$~2$) equal to zero. Then for the new sequence  $\sum_iip'_i$ diverges while $\sum_ip'_i$ is unchanged. With a similar argument you can see that you can also (with just slightly more effort) make $\sum_i(\ln i)p_i$ diverge, or $\sum_i(\ln(\ln i))p_i$, or anything similar.
A: I'm afraid it is infinity. 
Since it's been known that $ \sum\nolimits_{n = 1}^\infty  {\frac{6}{{n^2{\pi ^2}}}}  = 1 $, we could assume that $ p_i = \frac{6}{(i+1)^2\pi^2}$. Hence
\begin{eqnarray}
&&1+\sum\nolimits_{n = 1}^\infty  {np_n} \\
&=& 1 + \sum\nolimits_{n = 1}^\infty  {\frac{6n}{(n+1)^2\pi^2}} \\
&=& 1 + \frac{6}{\pi^2}\sum\nolimits_{n = 1}^\infty \left[ {\frac{1}{n+1}-\frac{1}{(n+1)^2}} \right] 
\end{eqnarray}
It seems that it is not convergent.
A: I think the case where $p_i$ = $p_0 \alpha^i$, $|\alpha| < 1$ is helpful to consider.
$$\sum_{i=0}^\infty p_i = p_0 \sum_{i = 0}^\infty \alpha^i,$$
$$ = \frac{p_0}{1 - \alpha},$$
so the normalization condition gives,
$$p_0 = 1 - \alpha.$$
Now for the second sum:
$$1 + p_1 + 2 p_2 + ... = 1 + \sum_{i = 1} i \left(p_0 \alpha^i\right),$$
$$= 1 + p_0 \frac{\alpha}{(1 - \alpha)^2},$$
$$= 1 + \frac{\alpha}{1 - \alpha}.$$
So to get a particular value for the sum, just pick $\alpha$.
