How to find the closed form for $\sum_{i=1}^k 3^{-i}$

Question

How do I calculate (is closed form the correct word?) the closed form of: $$\sum_{i=1}^k 3^{-i}?$$ I know how to calculate the geometric series: $$\sum_{i=1}^\infty 3^{-i} = \frac1{1-(1/3)}$$ But that doesn't help since I don't know how to calculate $\sum_{i=k+1}^\infty 3^{-i}$ (without calculating the thing I want). Wolframalpha says it is $$\sum_{i=1}^k 3^{-i}=\frac12 - \frac{3^{-k}}{2}$$ But I don't know how they calculated this. Maybe there is a closed form for $$\sum_{i=1}^k n^{-i},$$ for $n\in \Bbb N$?

Answer 1

$S_k = (1/3)^1+(1/3)^2+....+(1/3)^k$;

$(1/3)S_k= (1/3)^2+.....+(1/3)^{k+1}$;

Subtract:

$S_k-(1/3)S_k=$

$(2/3)S_k=(1/3)^1-(1/3)^{k+1}$;

$S_k=(3/2)((1/3)-(1/3)(1/3)^k)$;

$S_k =(1/2)-(1/2)(1/3)^k$;

$S_k= (1/2)-(1/2)3^{-k}.$

Answer 2

Hint: Use the well-known formula $$\sum_{i=1}^n q^i=\frac{q^{n+1}-q}{q-1}$$ for $q\ne 1$

Answer 3

If you know about the infinite geometric series sum (for $-1\le r<1$)

$$\sum_{i=1}^{\infty} r^i = \frac{r}{1-r}$$

Then a finite sum is just

$$\begin{align*} \sum_{i=1}^{k} r^i &= \sum_{i=1}^{\infty} r^i - \sum_{i=k+1}^{\infty} r^i\\ &= \sum_{i=1}^{\infty} r^i - r^k\sum_{i=1}^{\infty} r^i\\ &= \frac{r(1-r^k)}{1-r} \end{align*}$$

(This formula works for all $r\ne 1$, but this proof only works for $-1\le r<1$.)

Answer 4

we have: $$S_k=\sum_{i=1}^k3^{-i}$$ note that the terms of a geometric series can be expressed as: $$a_k=ar^{k-1}$$ in this case, we have: $a=\frac 13$,$r=\frac 13$. We can now use the formula: $$S_k=a\frac{(1-r^k)}{1-r}$$ this should give you: $$S_k=\frac{1}{2}\left(1-3^{-k}\right)$$ It can also be derived as given below

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we have: $$S_k=\frac{1}{3}+\frac{1}{3^2}+...\frac{1}{3^k}$$ which can be factorised to: $$S_k=\frac{1}{3}\left(1+\frac{1}{3}+...\frac{1}{3^{k-1}}\right)$$ and notice that this is just the same as: $$S_k=\frac 13\left(1+S_{k-1}\right)$$ we also know that: $$S_k=S_{k-1}+a_k=S_{k-1}+\frac{1}{3^k}\Rightarrow S_{k-1}=S_k-\frac{1}{3^k}$$ substituting one equation into the other, we get: $$3S_k=1+S_k-\frac{1}{3^k}$$ $$2S_k=1-\frac{1}{3^k}\Rightarrow S_k=\frac 12\left(1-3^{-k}\right)$$