Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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 at 22:34
1  
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 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 at 22:40
    
I can't follow you on this one Gottfried Helms! What exactly do you mean? –  hackYou Mar 15 at 22:43
    
With the revised formula, I now get the same result for both. –  celtschk Mar 15 at 22:45

4 Answers 4

up vote 2 down vote accepted

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.

share|improve this answer
    
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 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.

share|improve this answer
    
Thanks for the input. :) –  hackYou Mar 16 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 $\;n=1,2\;$ !

share|improve this answer
    
Looks like there was a typo in the original question, since corrected. –  Ben Millwood Mar 15 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 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 $\sin(n \pi /2)^2$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.