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

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.

• You (might) know that $\sum_{k=0}^{\infty} 1/2^k = 2$, so just look that the "negative part" of the series. Jun 23, 2020 at 21:08
• Amazing to see how most respondents want to keep the whole series, when a single term is enough to show divergence !
– user65203
Jun 23, 2020 at 21:17
• @Viktor Glombik, Yes.. Actually this is one of the previous years' question of my college for the current semester. So was looking for an exact solution. I tried something like this: $$\sum_{k=-n}^n 2^{-k}=2(2^n-1) +(2-2^{-n}).$$ Here the first part surely diverges as $n\to\infty.$ So $L=\infty.$ But wasn't very satisfied with this approach. So posted it here to see some elegant answers or suggestions. Thank you. Jun 23, 2020 at 21:19

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

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

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 $$L=\infty$$.

• Yes.. This was my approach. But was doubtful whether it would work or not. Because this question is one of the previous year question of my college. And our professor deducts our marks if he doesn't find our answers exactly to the point or if he is not satisfied with it. Thanks. Jun 23, 2020 at 21:24

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.

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

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

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

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

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