# Why is my series wrong?

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.

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When inserting $n=1$, your sequence gives $(3^0-7/2(-1)^1+1)/3^1 = (2 + 7/2)/3 = 11/6$, which clearly is different from $1/3$, while the professor's series correctly gives $(3-(-1)^2)/(2\cdot 3^1) = 2/6 = 1/3$. – celtschk Mar 15 '14 at 22:34
Hmm. If the general term is $$1/3+2/9+1/27+2/81+1/243+2/729$$ then the series seems to be $\\$ $$(1/3+2/9+1/27+2/81+1/243+2/729) + (1/3+2/9+1/27+2/81+1/243+2/729)+(1/3+2/9+1/27+2/81+1/243+2/729) + ...$$ $\\$ which I would assume to be -for some possible solution of this divergent case- $(1/3+2/9+1/27+2/81+1/243+2/729)\zeta(0)$. What do you think? – Gottfried Helms Mar 15 '14 at 22:39
Thank you for your effort celtschk and DonAntonio and I apologize becuase I made a mistake writing it into mathcad since i didn't know how properly write equations here. If you could take a look at it again please? – hackYou Mar 15 '14 at 22:40
I can't follow you on this one Gottfried Helms! What exactly do you mean? – hackYou Mar 15 '14 at 22:43
With the revised formula, I now get the same result for both. – celtschk Mar 15 '14 at 22:45

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.

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Thank you Ben. Looks like I'm just gonna have to tell him that your sln. looks nicer and that's it. – hackYou Mar 15 '14 at 23:15

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.

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Thanks for the input. :) – hackYou Mar 16 '14 at 16:39

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

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

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Looks like there was a typo in the original question, since corrected. – Ben Millwood Mar 15 '14 at 23:03
Yes @BenMillwood, it was edited: that denominator $\;2\;$ in the numerator only applied to $\;7(-1)^n\;$ ...Here is the original: i.stack.imgur.com/LO8AD.png – DonAntonio Mar 16 '14 at 0:11

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 $\sin(n \pi /2)^2$

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