What is the sum as n approaches infinity of $\displaystyle 5(\frac{1}{3})^{r - 1}$? Sorry I don't know how to write the Summation sign with the infinity symbol at the top and the r = 1 at the bottom of the summation sign in LaTeX.
What is the sum as n approaches infinity of $\displaystyle 5(\frac{1}{3})^{r - 1}$? Sorry I don't know how to write the Summation sign with the infinity symbol at the top and the r = 1 at the bottom of the summation sign in LaTeX.
Hello mark1950You want to know the answer to the summation $\displaystyle S=\sum_{r=1}^{\infty}5\times\Big(\frac{1}{3}\Big)^ {r-1}$.
Let's put the first few values of $\displaystyle r$ in:
$\displaystyle S = 5\times\Big(\frac{1}{3}\Big)^{0}+5\times\Big(\frac {1}{3}\Big)^{1}+5\times\Big(\frac{1}{3}\Big)^{2}+. ..$
$\displaystyle = 5 + 5\times\Big(\frac{1}{3}\Big)^{1}+5\times\Big(\frac {1}{3}\Big)^{2}+...$
which you may now recognise as a Geometric Progression (or Series) whose first term, $\displaystyle a= 5$, and common ratio $\displaystyle r=\frac{1}{3}$. The sum to infinity of such a series is $\displaystyle \frac{a}{1-r}$. So in this case:
$\displaystyle S = \frac{5}{1-\frac13}=\frac{5}{\tfrac23}=\frac{15}{2}=7.5$
Grandad
How about if the progressions aren't geometric? Like,
$\displaystyle \sum_{r=1}^{\infty}\frac{1}{(r + 1)(r + 2)}$
and
$\displaystyle \sum_{r=1}^{\infty}(r + 1)(r + 2^r)$ from the progressions, 2.3 + 3.6 + 4.11 + 5.64 + ...
Hello mark1950There are no hard and fast rules, because there are so many variations. But in this case, you use Partial Fractions, and say that
$\displaystyle \frac{1}{(r+1)(r+2)}= \frac{1}{r+1}-\frac{1}{r+2}$
so that
$\displaystyle \sum_{r=1}^{\infty}\frac{1}{(r + 1)(r + 2)}=\sum_{r=1}^{\infty}\Big(\frac{1}{r + 1}-\frac{1}{r + 2}\Big)$
$\displaystyle =\sum_{r=1}^{\infty}\frac{1}{r + 1}-\sum_{r=1}^{\infty}\frac{1}{r + 2}$
$\displaystyle =\sum_{r=1}^{\infty}\frac{1}{r + 1}-\sum_{r=2}^{\infty}\frac{1}{r + 1}$
$\displaystyle =\sum_{r=1}^{1}\frac{1}{r + 1}+\sum_{r=2}^{\infty}\frac{1}{r + 1}-\sum_{r=2}^{\infty}\frac{1}{r + 1}$
$\displaystyle =\frac12$
Grandad
Hi
I think you can do it using the method of differences but it is longer than Grandad's solution .
Same thing , do the partial fraction first .
$\displaystyle =\sum^n_{r=1}[\frac{1}{r+1}-\frac{1}{r+2}]$
$\displaystyle =\sum^n_{r=1}[f(r)-f(r+1)]$
$\displaystyle =f(1)-f(n+1)$
$\displaystyle =\frac{1}{2}-\frac{1}{n+2}$
$\displaystyle =\frac{n+1}{2n+4}$
Then $\displaystyle \sum_{r=1}^{\infty}\frac{1}{(r + 1)(r + 2)}=\lim_{n\rightarrow\infty}\frac{n+1}{2n+4}$
$\displaystyle =\frac{\frac{n}{n}+\frac{1}{n}}{\frac{2n}{n}+\frac {4}{n}}$
=1/2
Hello mark1950That is essentially the same as the method I've used here. It's just that instead of writing individual terms like:
$\displaystyle \frac12-\frac13+\frac13-\frac14+\frac14-\frac15+...$
I've left the series as two summations using the $\displaystyle \Sigma$ sign, and fiddled the limits around. It's just a bit neater that way.
Grandad