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- Homework Statement
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- Relevant Equations
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Using the given rule for the ##x_n##, write $$ \sum_l y_l = x_1 + \frac{1}{2} x_2 + \frac{1}{6} x_2 + \frac{1}{3} x_3 + \frac{1}{12} x_2 + \frac{1}{12} x_3 + \frac{1}{20} x_2 + \cdots + \frac{1}{n} x_n $$

$$ = x_1 + \sum_{n=2}^\infty \frac{1}{n(n-1)} x_2 + \sum_{n=3}^\infty \frac{2}{n(n-1)} x_3 + \sum_{n=4}^\infty \frac{3}{n(n-1)} x_4 + \cdots + \sum_{k=n}^\infty \frac{n-1}{k(k-1)} x_n $$

in general, ## \sum_{n=1}^m \frac{1}{n(n-1)} = \sum_l ( \frac{1}{n} - \frac{1}{n+1} ) = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} + \cdots \frac{1}{m-1} - \frac{1}{m} - \frac{1}{m+1} ## The sum collapses like a telescope and everything is canceled out except for the first and last terms. Its limit as ## m \longrightarrow \infty ## is ##1##. So,

$$ = x_1 + x_2 + x_3 + \cdots + x_n = \sum_{k=1}^n x_k$$

The infinite series ## \sum_{n=1}^\infty x_n = 1 - 1 + 1 - 1 + \cdots ## diverges since its value alternates between 0 and 1. But, we write ## \sum_n x_n = \sum_l y_l ## and assume the new series converges to a number ##S##.

##\sum_l y_1 = S = 1 + 1 - 1 + 1 + \cdots##

subtract ##1## from both sides

##S - 1 = -1 + 1 + 1 - + \cdots##

##S -1 = - S##

##S = \frac{1}{2}##

i'm unsure about the part where ##S## is used as a variable for a mere algebra trick. it seems like the result from the first part of the question isn't really used.

$$ = x_1 + \sum_{n=2}^\infty \frac{1}{n(n-1)} x_2 + \sum_{n=3}^\infty \frac{2}{n(n-1)} x_3 + \sum_{n=4}^\infty \frac{3}{n(n-1)} x_4 + \cdots + \sum_{k=n}^\infty \frac{n-1}{k(k-1)} x_n $$

in general, ## \sum_{n=1}^m \frac{1}{n(n-1)} = \sum_l ( \frac{1}{n} - \frac{1}{n+1} ) = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} + \cdots \frac{1}{m-1} - \frac{1}{m} - \frac{1}{m+1} ## The sum collapses like a telescope and everything is canceled out except for the first and last terms. Its limit as ## m \longrightarrow \infty ## is ##1##. So,

$$ = x_1 + x_2 + x_3 + \cdots + x_n = \sum_{k=1}^n x_k$$

The infinite series ## \sum_{n=1}^\infty x_n = 1 - 1 + 1 - 1 + \cdots ## diverges since its value alternates between 0 and 1. But, we write ## \sum_n x_n = \sum_l y_l ## and assume the new series converges to a number ##S##.

##\sum_l y_1 = S = 1 + 1 - 1 + 1 + \cdots##

subtract ##1## from both sides

##S - 1 = -1 + 1 + 1 - + \cdots##

##S -1 = - S##

##S = \frac{1}{2}##

i'm unsure about the part where ##S## is used as a variable for a mere algebra trick. it seems like the result from the first part of the question isn't really used.