Why is my series wrong?

Question

Why is this series wrong and how does it differ from this other one?

We had to find the general term for the series: $ 1/3+2/9+1/27+2/81+1/243+2/729+\ldots $ where the index begins at $n=1$ So I came up with this (see image, first formmula) now the profsaid this isn't right and gave us the sln.(see image, second one), so the next time I have to explain why this is wrong. .

\begin{align} \mathrm{an_{me}} &= \frac{3^{1+(-1)^n} - \frac{7}{2}[1+(-1)^n]}{3^n} \\ \\ \mathrm{an_{prof}} &= \frac{3-(-1)^{n+1}}{2\cdot 3^n} \end{align}

Again where do these series differ? I can't see any difference besides that my formula is kinda messy!

Thnx.

Answer 1

First, let's multiply up to make the denominators match:

\[\frac{3^{1+(-1)^n} - \frac{7}{2}[1+(-1)^n]}{3^n} = \frac{2\cdot 3^{1+(-1)^n} - 7[1+(-1)^n]}{2\cdot 3^n}\]

so, we now only need to check if the numerators match.

Since $n$ is only used in the numerator as an exponent of the base $-1$, it's sufficient to check the even case and the odd case are the same:

$n$ even: $2\cdot 3^{1+(-1)^n} - 7[1+(-1)^n] = 2\cdot 3^2 - 14 = 4$, while $3-(-1)^{n+1} = 4$.

$n$ odd: $2\cdot 3^{1+(-1)^n} - 7[1+(-1)^n] = 2\cdot 3^0 - 0 = 2$, while $3-(-1)^{n+1} = 2$.

So the two expressions are always equal.

Answer 2

$$\frac{2\cdot 3^{1+(-1)^n}-7(-1)^n+2}{2\cdot 3^n}=\begin{cases}\frac{2+7+2}{2\cdot3^n}=\frac{11}{2\cdot3^n}&,\;\;n\;\;\text{odd}\\{}\\\frac{2\cdot 9-7+2}{2\cdot3^n}=\frac{13}{2\cdot3^n}&,\;\;n\;\;\text{even}\end{cases}$$

It isn't anything close to the other thing, not even for $\;n=1,2\;$ !

Answer 3

The general term in terms of $n$ for $n=0 .. \infty$ can be expressed by many forms. One is $$ { 3 - (-1)^n \over 2 } \cdot {1 \over 3^{n+1}} $$ Another one is $$ (1+ \sin({n\pi /2 })^2 ) {1 \over 3^{n+1}} $$ The second form has the advantage, that it can be interpolated to any fractional index $n$ which might be meaningful in some contexts.
Other possibilities are to use any arbitrary one-periodic function giving $0,1,0,1,...$ at consecutive indexes instead of $\sin(n \pi /2)^2$

Answer 4

This is not really an answer, but a comment on a related issue you might find useful -- Millwood already gave a good answer. You can derive the professor's formula -- well, actually a simpler version as you can get rid of $n+1$ in the exponent and replace it with $n$ by a sign change -- as follows:

1. Recognize that $1, -1, 1, -1, ...$ is given by $(-1)^n$ (starting at $n = 0$)
2. Add $1$ to this to get $2, 0, 2, 0, 2, 0, ... $.
3. Divide this by $2$ to get $1, 0, 1, 0, 1, 0, ... $.
4. Add $1$ to that to get $2, 1, 2, 1, 2, 1, ... $.
5. You now have $\frac{(-1)^n + 1}{2} + 1$.
6. Starting at $n = 1$, this is $1, 2, 1, 2, 1, ... $.
7. Just divide by $3^n$ and simplify.
8. To get the original form: multiply and divide $(-1)^n$ by $-1$ to get $n+1$ in the exponent and it as negative.

Done.