Are these two summations equivalent?

Question

Is $\sum_{y=2}^{\infty} (\frac{1}{y})(1-p)^{y-1}$ equivalent to $\sum_{y=1}^{\infty} (\frac{1}{y})(1-p)^{y}$ ?

Answer 1

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 $p = 1$.

Answer 2

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)