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$$\sum_{n=0}^{\infty}\frac 12 \left(\frac14\right)^{n-1}.$$

This is what I tried to do $$\frac 12\sum_{n=0}^\infty \left(\frac14\right)^{n-1} =\frac 12 \sum_{n=0}^\infty \left(-\frac14\right) \left(\frac14\right)^n =-\frac 18\sum_{n=0}^\infty \left(\frac14\right)^n=-\frac 18\cdot \frac1{1-\frac 14}.$$ Final answer: $\frac{-1} 6$

However, when I use online summation calculators, $2.66667$ is the answer.

What am I doing wrong?

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    $\begingroup$ It's because $(1/4)^{n-1}$ does not simplify to $(-1/4)(1/4)^n$ $\endgroup$ – rb612 May 24 '17 at 1:31
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Your second step is flawed, $a^{n-1} \neq (-a)a^n$.

We have $a^{n-1} = \frac{1}{a}a^n$, so your second step should read:

$$\frac{1}{2} \sum_{n=0}^\infty4 \left(\frac{1}{4}\right)^n$$

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You need to revise power laws: $$\Bigl(\frac14\Bigr)^{n-1} =\Bigl(\frac14\Bigr)^{n}\Bigl(\frac14\Bigr)^{-1} =4\Bigl(\frac14\Bigr)^{n}\ ,$$ not $$\Bigl(-\frac14\Bigr)\Bigl(\frac14\Bigr)^{n}$$ as you wrote.

BTW it should be obvious that your answer is wrong, because you have a series of positive numbers adding up to a negative value.

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It is a Geometric series. You can find properties of this kind of series here linked-to result.

For the case in your question we have:

$\sum_{n=0}^{\infty} (\frac{1}{2}) (\frac{1}{4})^{n-1} = \frac{1}{2} \times \frac{1}{4}^{-1} \sum_{n=0}^{\infty}(\frac{1}{4})^{n} = 2\times \frac{1}{1 - \frac{1}{4}} = 2 \times \frac{4}{3}= \frac{8}{3}$.

And we are done.

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