Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/237

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 * 10. $$\frac{n}{2^n}$$. || $$\frac{1}{2},\, \frac{2}{4},\, \frac{3}{8},\, \frac{4}{16}$$.
 * 11. $$\frac{\left ( \log a \right )^nx^n}{n!}$$ || $$\frac{\log a \cdot x}{1},\, \frac{\log^2 a \cdot x^2}{2},\, \frac{\log^3 a \cdot x^3}{6},\, \frac{\log^4 a \cdot x^4}{24}$$.
 * 12. $$\frac{\left ( -1 \right )^{n-1}x^{2n-2}}{\left ( 2n - 1 \right ) !}$$. || $$\frac{1}{1},\, -\frac{x^2}{3!},\, \frac{x^4}{5!},\, -\frac{x^6}{7!}$$.
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 * 12. $$\frac{\left ( -1 \right )^{n-1}x^{2n-2}}{\left ( 2n - 1 \right ) !}$$. || $$\frac{1}{1},\, -\frac{x^2}{3!},\, \frac{x^4}{5!},\, -\frac{x^6}{7!}$$.
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Consider the series of $$n$$ terms

and let $$S_n$$ denote the sum of the series. Then

Evidently $$S_n$$ is a function of $$n$$, for

$$ \begin{array}{lcllclcl} \text{when}\ n & = & 1, & S_1 & = & 1                 & = & 1, \\ \text{when}\ n & = & 2, & S_2 & = & 1 + \frac{1}{2}   & = & 1\frac{1}{2}, \\ \text{when}\ n & = & 3, & S_3 & = & 1 + \frac{1}{2} + \frac{1}{4} & = & 1\frac{3}{4}, \\ \text{when}\ n & = & 4, & S_4 & = & 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} & = & 1\frac{7}{8}, \\ &  &    &     &   &    \cdots                                   &   & \\ \text{when}\ n & = & n, & S_n & = & 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots + \frac{1}{2^{n-1}} & = & 2 - \frac{1}{2^{n-1}}, \end{array} $$

Mark off points on a straight line whose distances from a fixed point correspond to these different sums. It is seen that the point

corresponding to any sum bisects the distance between the preceding point and 2. Hence it appears geometrically that when $$n$$ increases without limit

$$\lim S_n = 2$$.

We also see that this is so from arithmetical considerations, for $$\lim_{n \to \infty}S_n = lim_{n \to \infty} \left ( 2 - \frac{1}{2^{n-1}} \right ) = 2$$.

[Since when $$n$$ increases without limit $$\tfrac{1}{2^{n-1}}$$ approaches zero as a limit.]