Find $\sum_{n=-\infty}^{\infty} (0.5)^n$

Question

We have: $$L=\lim_{n\to\infty} \sum_{k=-n}^n \frac1{2^k}.$$ The limit surely diverges or tends to $\infty.$ But I can't think of a proper way to show this.

Please suggest, how can I show that $L=\infty\,?$ Thanks in advance.

Answer 1

$$\sum_{k=-n}^n\frac1{2^k}\ge2^n$$ just by the term $k=-n$.

Answer 2

Hint: Note that $$ \sum_{k=-n}^n \frac1{2^k} = 2^n \cdot \sum_{k=0}^{2n}\frac 1{2^k} $$

Answer 3

Let's split up the sum as follows:

$$L=\lim_{n\to\infty}\left(\sum_{k=-n}^{-1}\dfrac{1}{2^k}+\sum_{k=0}^n\dfrac{1}{2^k}\right)$$

We notice that $\sum_{k=-n}^{-1}\frac{1}{2^k}=\sum_{k=1}^n\frac{1}{2^{-k}}$. Now we put this back into the expression for $L$ to get:

$$L=\lim_{n\to\infty}\left(\sum_{k=1}^n 2^k+\sum_{k=0}^n\dfrac{1}{2^k}\right)$$

From here, it's clear that the left sum diverges, and so $L=\infty$.

Answer 4

Split your sum in half: $$\sum_{n=-\infty}^\infty 2^{-n}=\sum_{n=-\infty}^0 2^{-n}+\sum_{n=1}^\infty 2^{-n}$$

Changing the base of the first sum by using $m=-n$, we have that:

$$\sum_{n=-\infty}^0 2^{-n}=\sum_{m=\infty}^0 2^{m}=\sum_{m=0}^\infty2^m=\infty$$ which is easily proven by any useful test.

Answer 5

$$\sum_{k=-n}^n\frac{1}{2^k}=$$

$$2^n+2^{n-1}+...2+1+\frac 12+...+\frac{1}{2^n}=$$

$$2^n\Bigl(1+\frac 12+...+\frac{1}{2^{2n}}\Bigr)=$$

$$2^n\Bigl(\frac{1-\frac{1}{2^{2n+1}}}{1-\frac 12}\Bigr)=$$

$$2^{n+1}(1-\frac{1}{2^{2n+1}})$$ the limit is $$L=+\infty \times 1=+\infty$$