Finding what the sum of $(2^n)(3^{(-n+1)})$ from $n=1$ to infinity converges to

Question

Sorry for the lack of math symbols but I'm trying to find what the sum of $f(x)=(2^n)(3^{-n+1})$ from $n=1$ to infinity converges to. I have tried using the sequence of partial sums but I'm too stupid to find what Sn is equal to in order to take the limit. Any help would be greatly appreciated.

Answer 1

Define $$S=\sum_{n=1}^{\infty}2^n\cdot3^{-n+1}.$$

Observe that $$S=\sum_{n=1}^{\infty}2^n\cdot3^{-n+1}=-2^0\cdot3^{-0+1}+\sum_{n=0}^{\infty}2^n\cdot3^{-n+1}=-3+\underbrace{\sum_{n=0}^{\infty}2^n\cdot3^{-n+1}}_{:=S_0}.$$ Rewrite the summand: $$S_0=3\sum_{n=0}^{\infty} \frac{2^n}{3^{n}}=3\sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^n.$$ By the geometric series this equal to $$\frac{3}{1-\frac{2}{3}}=9.$$ Since $S=S_0-3$, the final answer is $S=6$.

Answer 2

$2^n3^{-n+1} = 2^n3^{-(n-1)} =2^n\cdot \frac 1{3^{n-1}} = 2^n\cdot \frac {3}{3^{n}} = 3\cdot \frac {2^n}{3^n} = 3(\frac 23)^n$.

So we need $\sum_{n=1}^{\infty} 3\cdot (\frac 23)^n$.

Can you take from there?

Well known formula: If $|r| < r$ then $\sum_{n=\color{green}0}^{\infty} r^n = \frac 1{1-r}$. (Note the base step is $n = \large{\color{green}0}$ and not $n=\large{\color{red}1}$).

Derivation if you didn't know the above result.

$(1 + r + r^2 + r^3 + ........+r^m)(1-r) = $
$(1 + r + r^2 + r^3 + ........+r^m) - (r+ r^2 + r^3 + r^4 + ...... + r^{m+1})=$
$1 - r^{m+1}$. So

$ (1 + r + r^2 + r^3 + ........+r^m) =\frac {1-r^{m+1}}{1-r}$

if $|r| < 1$ then $\lim_{m\to \infty} r^{m+1} = 0$ so

$\sum_{n=0}^\infty r^k =\lim_{m\to\infty} \sum_{n=0}^m r^k = \lim_{m\to \infty}\frac {1-r^m}{1-r} = \frac 1{1-r}$.

Actual Answer (but only read if you really, really, really need it!)

$\sum_{n=1}^{\infty} 3\cdot (\frac 23)^n =$
$3 \sum_{n=1}^{\infty} (\frac 23)^n = $
$3([\sum_{n=\color{red}{\large 0}}^{\infty}(\frac 23)^n]- (\frac 23)^{\color{red}{\large 0}}) =$
$3(\frac 1{1-\frac 23} - 1)=3(\frac 1{\frac 13} - 1)=3(3-1)=3\cdot 2 = 6$

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Might be useful to try to calculate

$2^1\cdot 3^{-1+1} + 2^2 \cdot 3^{-2+1} + .......... + 2^m \cdot 3^{-m+1}$ for a finite $m$.

$2^1\cdot 3^{-1+1} + 2^2 \cdot 3^{-2+1} + .......... + 2^m \cdot 3^{-m+1}=$
$(2^1\cdot 3^{-1+1} + 2^2 \cdot 3^{-2+1} + .......... + 2^m \cdot 3^{-m+1} )\frac {1-\frac 23}{1-\frac 23} =$ $\frac {(2^1\cdot 3^{-1+1} + 2^2 \cdot 3^{-2+1} + .......... + 2^m \cdot 3^{-m+1} )(1- \frac 23)}{\frac 13} =$ $3\cdot (2^1\cdot 3^{-1+1} + 2^2 \cdot 3^{-2+1} + .......... + 2^m \cdot 3^{-m+1} )(1- \frac 23)=$ $3[(2^1\cdot 3^{-1+1} + 2^2 \cdot 3^{-2+1} + .......... + 2^m \cdot 3^{-m+1} )- (2^2 \cdot 3^{-2+1} + 2^3 \cdot 3^{-3+1} + ........ + 2^{m+1}\cdot 3^{-m})]=$ $3[2^1 \cdot 3^{-1+1} - 2^{m+1}\cdot 3^{-m})]=$ $3[2 - 2\cdot \frac {2^m}{3^m}]=$ $6 - 6(\frac 23)^m$.
As $6(\frac 23)^m \to 0$ we get the infinite sum is $6$.