The first 3 terms of a series are given and the sum to n terms equation. Can someone explain what this technique does? It obviously divides each factor with n in it by n.
Hello, Stuck Man!
$\displaystyle \text{Given: the first 3 terms of a series and the sum to }n\text{ terms formula.}$
$\displaystyle \text{Can someone explain what this technique does?}$
What is your question?
Do you want their solution explained?
$\displaystyle \displaystyle}\frac{5}{2\cdot3\cdot4} + \frac{7}{3\cdot4\cdot5} + \frac{9}{4\cdot5\cdot6} + \hdots$
$\displaystyle S_n \;=\;\dfrac{(2n+3)(2n+5)}{2(n+2)(n+3)} - \dfrac{5}{4}$
$\displaystyle \text{Find the sum of the series to infinity.}$
$\displaystyle \text{Write }S_n \text{ as: }\;\displaystyle \frac{(2+\frac{3}{n})(2+\frac{5}{n})}{2(1+\frac{2} {n})(1+\frac{3}{n})} - \frac{4}{5}$
$\displaystyle \text{Show that: }\;\displaystyle{\lim_{n\to\infty} S_n \;=\;\frac{3}{4}$
They gave us this formula: .$\displaystyle S_n \;=\;\dfrac{(2n+3)(2n+5)}{2(n+2)(n+3)} $
Divide numerator and denominator by $\displaystyle n^2\!:$
. . $\displaystyle \displaystyle S_n \;=\; \frac{(\frac{2n+3}{n})(\frac{2n+5}{n})}{2(\frac{n+ 2}{n})(\frac{n+3}{n})}-\frac{4}{5} \;=\; \frac{(2+\frac{3}{n})(2+\frac{5}{n})}{2(1+\frac{2} {n})(1+\frac{3}{n})} - \frac{4}{5} $
Take the limit:
. . $\displaystyle \displaystyle \lim_{n\to\infty} S_n \;=\;\lim_{n\to\infty}\frac{(2+\frac{3}{n})(2+\fra c{5}{n})}{2(1+\frac{2}{n})(1+\frac{3}{n})} - \frac{5}{4}$
. . . . . . . . $\displaystyle \displaystyle =\; \frac{(2+0)(2+0)}{2(1+0)(1+0)} - \frac{5}{4} \;\;=\;\;\frac{4}{2} - \frac{5}{4} \;\;=\;\;\frac{3}{4}$