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1) In my guide I have that:

$$ \sum_{k=1}^{\infty }p_{k}=\sum_{k=1}^{\infty }\frac{1}{3}\times 2^{-2k+1} = 2/9$$

I can't figure out what I have this.

2) Plus, the original probability function was $$ p_{k}=\frac{1}{3}\times 2^{-\begin{vmatrix} k\end{vmatrix}}$$ The guide says that I need to evaluate for $$p_{-2k+1}$$

So the answer provided by the guide is not correct either?

Thanks.

Norse
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Limanido
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2 Answers2

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Using Geometric Series sum

$$\sum_{k=1}^\infty2^{1-2k} =\dfrac{\dfrac12}{1-\dfrac1{2^2}}=?$$

lab bhattacharjee
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Do you not realize that, with $k\ge 1$, $-2k+ 1\le -1$ so that |-2k+1|= 2k- 1? So your sum is $\sum_{k=1}^\infty \frac{1}{3} 2^{2k-1}= \frac{1}{6}\sum 4^k$. That sum clearly does not converge!

However, without the absolute value, $2^{-2k+ 1}= 2(4^{-k})$ so you have $\frac{1}{3}\sum_{k=1}^\infty 2^{-2k+ 1}= \frac{2}{3}\sum_{k=1}^\infty \left(\frac{1}{4}\right)$ and that last sum is a geometric series which has a simple formula. $\frac{1}{3}\sum_{k=1}^\infty 2^{-2k+ 1}= \frac{2}{3}\frac{1}{1-\frac{1}{4}}= \frac{2}{3}\frac{4}{3}= \frac{8}{9}$.

user247327
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  • The summation runs from $k=1$, not $k=0$, so your answer differs from the expected answer by a factor of $4$ - i.e., you have a 'matho'. Also, FWIW, the OP's definition of $p_k$ had a minus sign outside the absolute value in the exponent, so probably the OP's original guide did not have a typo, after all. – peter a g Dec 19 '19 at 13:41
  • Actually, rereading the question, my 'probably' in my second comment was unwarranted: let's replace it with 'possibly'. – peter a g Dec 19 '19 at 14:00