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Find the sum of the following geometric series $$\sum_{k=2}^{\infty} \frac{5}{2^k} = \frac{5}{2}$$

Attempt:

First I test with the root criterion if its divergent or convergent... $\frac {1}{2} < 1$ so it's convergent...

Now I try to find the sum and I can't get to the solution $\frac{5}{2}$.

Let's try:

$$\sum_{k=2}^{\infty} \frac{5}{2^k}=5\sum_{k=2}^{\infty} \frac{1}{2^k}=5\sum_{k=2}^{\infty} 2^{-^k}=5\cdot2^{-2}\sum_{k=2}^{\infty}2^{-k}$$

I don't know how to work with the $2^{-k}$ in here, hope for your help^^

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    $\begingroup$ The last sum should start from $k=0$. But you don't need to go through all this. If $|r|<1$, then $$a+ar+ar^2+\cdots = \frac {a}{1-r}$$ $\endgroup$
    – saulspatz
    Mar 11, 2021 at 17:14
  • $\begingroup$ $\sum_{k=1}^{\infty} \frac{1}{2^k} = \frac{0.5}{1-0.5}=1$ should give you a hint. $\endgroup$
    – Talmsmen
    Mar 11, 2021 at 17:14
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    $\begingroup$ Typesetting hint: you don't need to surround every symbol with dollar signs; use them just at the beginning and end of math expressions. For example, instead of $2$$^-$$^k$, try $2^{-k}$. Please have a look at the typesetting tutorial and reference. $\endgroup$
    – Théophile
    Mar 11, 2021 at 17:17

3 Answers 3

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$$\sum_{k=2}^{\infty} \frac{1}{2^k} = \sum_{k=0}^{\infty} \frac{1}{2^k} - (1 +\frac{1}{2}) = \frac{1}{1 - \frac{1}{2}} -\frac{3}{2} = \frac{1}{2}$$So the answer is $5\times\frac{1}{2} = \frac{5}{2}$

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Actually,$$\sum_{k=2}^\infty2^{-k}=2^{-2}\sum_{k=0}^\infty2^{-k}=2^{-2}\times2=\frac12.$$

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You can combine the answers from S.H.W and José Carlos Santos to figure out a solution if you don't know the sum $\sum^\infty_{k=0} 2^{-k}$.

$$ \begin{align} \sum^\infty_{k=2} 2^{-k} &= \sum^\infty_{k=0}2^{-k} - 1 - \frac{1}{2} \tag{S.H.W} \\ &=2^{-2}\sum^\infty_{k=0}2^{-k} \tag{José Carlos Santos} \end{align} $$

If we write $L=\sum^\infty_{k=0}$ then $L-\frac{3}{2} = \frac{L}{4}$. This has as only solution $L=2$. Therefore: $$\sum^\infty_{k=2} = \frac{L}{4} = \frac{1}{2}$$

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