MHB 242.10.3.27 using the geometric formula of a sum

[karush]

$\tiny{242.10.3.27}$
evaluate
$$S_j=\sum_{j=1}^{\infty}3^{-3j}=$$
rewrite
$$S_j=\sum_{j=1}^{\infty} 27^{j-1}$$
using the geometric formula
$$\sum_{n=1}^{\infty}ar^{n-1}=\frac{a}{1-r}, \left| r \right|<1$$
how do we get $a$ and $r$ to get the answer of $\frac{1}{26}$

 

[kaliprasad]

$\tiny{242.10.3.27}$
evaluate
$$S_j=\sum_{j=1}^{\infty}3^{-3j}=$$
rewrite
$$S_j=\sum_{j=1}^{\infty} 27^{j-1}$$
using the geometric formula
$$\sum_{n=1}^{\infty}ar^{n-1}=\frac{a}{1-r}, \left| r \right|<1$$
how do we get $a$ and $r$ to get the answer of $\frac{1}{26}$

there are a couple of issues here
$S_j=\sum_{j=1}^{\infty}3^{-3j}=$
=> $S_j=\sum_{j=1}^{\infty} 27^{-j}$
$= \sum_{j=1}^{\infty} (\frac{1}{27})^j$
$= \frac{1}{27}\sum_{j=0}^{\infty} (\frac{1}{27})^j$ (kinldy note that the sum index is from 0 and not 1)
in the above $a= \frac{1}{27}$ and $r =\frac{1}{27}$
= $\frac{1}{27}( \frac{1}{1-\frac{1}{27}})= \frac{1}{26}$

 

[karush]

thanks that explained things...

 

[karush]

$\tiny{242.10.3.27}$
$\text{evaluate}\\$
$$S_j=\sum_{j=1}^{\infty}3^{-3j}=$$
$\text{rewrite} \\$
$$\displaystyle
S_j=\sum_{j=1}^{\infty} 27^{-j}
=\sum_{j=1}^{\infty}\left(\frac{1}{27}\right)^{j}
=\frac{1}{27}\sum_{j=0}^{\infty}\left(\frac{1}{27}\right)^{j}$$
$\text{using the geometric formula }$
$$\displaystyle
\sum_{n=1}^{\infty}ar^{n-1}
=\frac{a}{1-r}, \left| r \right|<1$$
$\text{then }$ $a=\frac{1}{27}$ $\text{ and } r=\frac{1}{27} \\$
$$S_j= \frac{1}{27}\left[\frac{1}{1-\frac{1}{27}}\right]=\frac{1}{26}$$

 

[kaliprasad]

$\tiny{242.10.3.27}$
$\text{evaluate}\\$
$$S_j=\sum_{j=1}^{\infty}3^{-3j}=$$
$\text{rewrite} \\$
$$\displaystyle
S_j=\sum_{j=1}^{\infty} 27^{-j}
=\sum_{j=1}^{\infty}\left(\frac{1}{27}\right)^{j}
=\frac{1}{27}\sum_{j=0}^{\infty}\left(\frac{1}{27}\right)^{j}$$
$\text{using the geometric formula }$
$$\displaystyle
\sum_{n=1}^{\infty}ar^{n-1}
=\frac{a}{1-r}, \left| r \right|<1$$
$\text{then }$ $a=\frac{1}{27}$ $\text{ and } r=\frac{1}{27} \\$
$$S_j= \frac{1}{27}\left[\frac{1}{1-\frac{1}{27}}\right]=\frac{1}{26}$$

above is right except that
$$\displaystyle
\sum_{n=1}^{\infty}ar^{n-1}
=\frac{a}{1-r}, \left| r \right|<1$$

should be
$$\displaystyle
\sum_{0}^{\infty}ar^{n-1}
=\frac{a}{1-r}, \left| r \right|<1$$

 

[karush]

above is right except that
$$\displaystyle
\sum_{n=1}^{\infty}ar^{n-1}
=\frac{a}{1-r}, \left| r \right|<1$$

should be
$$\displaystyle
\sum_{0}^{\infty}ar^{n-1}
=\frac{a}{1-r}, \left| r \right|<1$$

this is from the textbook

 

[MarkFL]

This is what W|A returns (for $|r|<1$):

$$\sum_{k=0}^{\infty}\left(ar^{k-1}\right)=\frac{a}{r(1-r)}$$

$$\sum_{k=1}^{\infty}\left(ar^{k-1}\right)=\frac{a}{1-r}$$

However, these imply:

$$\sum_{k=0}^{\infty}\left(ar^{k}\right)=\frac{a}{1-r}$$

$$\sum_{k=1}^{\infty}\left(ar^{k}\right)=\frac{ar}{1-r}$$

 

[HallsofIvy]

above is right except that
$$\displaystyle
\sum_{n=1}^{\infty}ar^{n-1}
=\frac{a}{1-r}, \left| r \right|<1$$

should be
$$\displaystyle
\sum_{0}^{\infty}ar^{n-1}
=\frac{a}{1-r}, \left| r \right|<1$$

No, that would be for r^nnot r^{n-1}.