Are these two summations equivalent? Is $\sum_{y=2}^{\infty} (\frac{1}{y})(1-p)^{y-1}$ equivalent to  $\sum_{y=1}^{\infty} (\frac{1}{y})(1-p)^{y}$ ?
 A: We'll go ahead and evaluate both the sums to make things more interesting. Recall that for $|x| < 1$,
$$\frac{1}{1-x} = 1 + x + x^2  + x^3 + ...$$
Integrating both sides, we get
$$-\ln(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + ...$$
Substitute $x =  1-p$ (I'm assuming $| 1- p| < 1$). Then,
$$-\ln p = (1 - p) + \frac{1}{2}(1-p)^2 + \frac{1}{3}(1-p)^3 + \frac{1}{4}(1-p)^4 + \dots$$
So it follows that:
$$B = \sum_{y = 1}^{\infty} \frac{1}{y}(1-p)^y = -\ln p$$
Moving on, we subtract $(1-p)$ from both sides of our earlier result. This gives us:
$$-\ln p - (1-p) = \frac{1}{2}(1-p)^2 + \frac{1}{3}(1-p)^3 + \frac{1}{4}(1-p)^4 +\frac{1}{5}(1-p)^5 + \dots$$
$$\frac{-\ln p - (1-p)}{1-p} = \frac{1}{2}(1-p) + \frac{1}{3}(1-p)^2 + \frac{1}{4}(1-p)^3 +\frac{1}{5}(1-p)^4 + \dots$$
So we have:
$$A = \sum_{y = 2}^{\infty} \frac{1}{y}(1-p)^{y-1} = \frac{-\ln p - (1-p)}{1-p}$$
The rest is trivial. If $A = B$, then
$$\frac{-\ln p - (1-p)}{1-p} = -\ln p$$
$$\frac{\ln p + (1-p)}{1-p} = \ln p$$
$$\ln p + (1-p) = (1 - p) \ln p$$
$$\ln p + 1 - p = \ln p - p\ln p$$
$$-p\ln p = 1 - p$$
whose only (real) solution to this is when $p = 1$ (why? I'll leave this to you).
Hence, the two are equal if and only if $p = 1$.
A: Let's call $S_n$ (limit=S) the partial sum of the right series, and $T_n$ (limit=T) the other:
Tn = $\sum_{y=1}^{n} (\frac{1}{y})(1-p)^{y} = (1-p)*{( 1 + \sum_{y=2}^{n} (\frac{1}{y})(1-p)^{y-1}})$
When n--> ∞ :
L = (1-p)*( 1 + T)
