Prove some formulas for given infinite sums

by RoRi

February 29, 2016

We say that series

$\sum_{n=1}^{\infty} a_n \qquad \text{and} \qquad \sum_{n=1}^{\infty} b_n$

are identical if $a_n = b_n$ for all $n \geq 1$ . Determine whether each of the following pairs of series are identical:

$1 - 1 + 1 - 1 + \cdots$ and $(2-1) - (3-2) + (4-3) - (5-4) + \cdots$ .
$1 - 1 + 1 - 1 + \cdots$ and $(1-1) + (1-1) + (1-1) + (1-1) + \cdots$ .
$1 - 1 + 1 - 1 + \cdots$ and $1 + (-1+1) + (-1+1) + (-1+1) + \cdots$ .
$1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots$ and $1 + \left(1 - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{8} \right)+ \cdots$ .

These are identical since $a_n = (-1)^{n-1}$ , and $b_n = (-1)^{n-1}(n+1-n) = (-1)^{n-1}$ .
These are not identical since $a_n = (-1)^{n-1}$ , but $b_n =0$ .
These are not identical since $a_n = (-1)^{n-1}$ , but $b_n = 0$ , $b_1 = 1$ .
These are identical since $a_n = \frac{1}{2^{n-1}}$ , and $b_n = \frac{1}{2^{n-1}} - \frac{1}{2^n} = \frac{1}{2^{n-1}}$ .

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