What is the sum as n approaches infinity?

**mark1950** · September 7th 2009, 04:56 AM

What is the sum as n approaches infinity of $5(\frac{1}{3})^{r - 1}$ ? 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.

**Grandad** · September 7th 2009, 05:46 AM

Hello mark1950

Originally Posted by mark1950

What is the sum as n approaches infinity of $5(\frac{1}{3})^{r - 1}$ ? 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.

You want to know the answer to the summation $S=\sum_{r=1}^{\infty}5\times\Big(\frac{1}{3}\Big)^ {r-1}$ .

Let's put the first few values of $r$ in:

$S = 5\times\Big(\frac{1}{3}\Big)^{0}+5\times\Big(\frac {1}{3}\Big)^{1}+5\times\Big(\frac{1}{3}\Big)^{2}+. ..$

$= 5 + 5\times\Big(\frac{1}{3}\Big)^{1}+5\times\Big(\frac {1}{3}\Big)^{2}+...$

which you may now recognise as a Geometric Progression (or Series) whose first term, $a= 5$ , and common ratio $r=\frac{1}{3}$ . The sum to infinity of such a series is $\frac{a}{1-r}$ . So in this case:

$S = \frac{5}{1-\frac13}=\frac{5}{\tfrac23}=\frac{15}{2}=7.5$

Grandad

**mark1950** · September 7th 2009, 05:53 AM

How about if the progressions aren't geometric? Like,

$\sum_{r=1}^{\infty}\frac{1}{(r + 1)(r + 2)}$

and

$\sum_{r=1}^{\infty}(r + 1)(r + 2^r)$ from the progressions, 2.3 + 3.6 + 4.11 + 5.64 + ...

**Grandad** · September 7th 2009, 06:14 AM

Hello mark1950

Originally Posted by mark1950

How about if the progressions aren't geometric? Like,

$\sum_{r=1}^{\infty}\frac{1}{(r + 1)(r + 2)}$

and you have to find the sum to infinity?

There are no hard and fast rules, because there are so many variations. But in this case, you use Partial Fractions, and say that

$\frac{1}{(r+1)(r+2)}= \frac{1}{r+1}-\frac{1}{r+2}$

so that

$\sum_{r=1}^{\infty}\frac{1}{(r + 1)(r + 2)}=\sum_{r=1}^{\infty}\Big(\frac{1}{r + 1}-\frac{1}{r + 2}\Big)$

$=\sum_{r=1}^{\infty}\frac{1}{r + 1}-\sum_{r=1}^{\infty}\frac{1}{r + 2}$

$=\sum_{r=1}^{\infty}\frac{1}{r + 1}-\sum_{r=2}^{\infty}\frac{1}{r + 1}$

$=\sum_{r=1}^{1}\frac{1}{r + 1}+\sum_{r=2}^{\infty}\frac{1}{r + 1}-\sum_{r=2}^{\infty}\frac{1}{r + 1}$

$=\frac12$

Grandad

**mark1950** · September 7th 2009, 06:18 AM

Is it possible if I use the method of difference? If I used it, how would it looked like? Sorry for my inquiry.

**mathaddict** · September 7th 2009, 07:34 AM

Originally Posted by mark1950

Is it possible if I use the method of difference? If I used it, how would it looked like? Sorry for my inquiry.

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 .

$=\sum^n_{r=1}[\frac{1}{r+1}-\frac{1}{r+2}]$

$=\sum^n_{r=1}[f(r)-f(r+1)]$

$=f(1)-f(n+1)$

$=\frac{1}{2}-\frac{1}{n+2}$

$=\frac{n+1}{2n+4}$

Then $\sum_{r=1}^{\infty}\frac{1}{(r + 1)(r + 2)}=\lim_{n\rightarrow\infty}\frac{n+1}{2n+4}$

$=\frac{\frac{n}{n}+\frac{1}{n}}{\frac{2n}{n}+\frac {4}{n}}$

=1/2

**Grandad** · September 7th 2009, 07:38 AM

Hello mark1950

Originally Posted by mark1950

Is it possible if I use the method of difference? If I used it, how would it looked like? Sorry for my inquiry.

That is essentially the same as the method I've used here. It's just that instead of writing individual terms like:

$\frac12-\frac13+\frac13-\frac14+\frac14-\frac15+...$

I've left the series as two summations using the $\Sigma$ sign, and fiddled the limits around. It's just a bit neater that way.

Grandad

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