This question might have been asked before but I have not been able to find it.
How can I find: $$\sum_{i=0}^n 2^{-i}$$ Help?
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Do you need $\sum_{i=0}^n 2^i$ or $\sum_{i=0}^n 2^{-i}$? Anyway, do you know about GP series? – StubbornAtom Sep 27 '17 at 06:08
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@123 sorry I mean 0 to n – Lucas Sep 27 '17 at 06:09
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@StubbornAtom the second one is what I am looking for ∑ni=02−i∑i=0n2−i – Lucas Sep 27 '17 at 06:10
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@StubbornAtom I just started reading about GP series – Lucas Sep 27 '17 at 06:10
4 Answers
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If you know what is different base representation, then $\sum\limits_{i=0}^{n}2^i$ will look like $11\ldots 11_2$ with $n+1$ 1s. And any power of 2 is 1 followed by 0s, like $10\ldots 00_2$ with number of 0s equal to the power $2^n = 1\underbrace{0\ldots 0_2}_{n}$.
So, $$\sum\limits_{i=0}^{n}2^i = 2^{n+1}-1$$
Edit: for the $\sum\limits_{i=0}^{n}2^{-i}$
$$\sum\limits_{i=0}^{\infty}2^{-i} = 2$$ and $$\sum\limits_{i=0}^{n}2^{-i} = \sum\limits_{i=0}^{\infty}2^{-i} - \sum\limits_{i=n+1}^{\infty}2^{-i}=\sum\limits_{i=0}^{\infty}2^{-i}-2^{-n-1}\cdot\sum\limits_{i=0}^{\infty}2^{-i} = 2 - 2^{-n-1}\cdot2 = 2 - 2^{-n}$$
dEmigOd
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@Lucas look at your question body before I edited. This answer is valid. – Kenny Lau Sep 27 '17 at 06:23
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Now that you know the answer you can convert the binary number: 1.1111111111....., n+1, digits are =1, all others 0, to a decimal number. – Peter Szilas Sep 27 '17 at 07:26
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You have to find $$S=\sum_{i=0}^n2^{i}$$. Check the value of $2S$. By subtracting you will see that $$2S-S=2^{n+1}-2^{0}$$
QED
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Consider the infinite sum. Observe that if you divide it by $2$, you keep the same sum, with the initial term dropped. Then from $\dfrac S2=S-1$ you draw $S=2$.
Then if you want to truncate the sum after the $n^{th}$ term, notice that the tail of the sum is $2^{-n}S$. The rest is yours.
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This summation is particularly easy if you spend the effort of computing a few terms.
$$1,\\1+\frac12=\frac32,\\1+\frac12+\frac14=\frac74,\\1+\frac12+\frac14+\frac18=\frac{15}8,\\\cdots$$
Now it should be obvious that by repeating the last term, you obtain $2$ in all cases.