Alternating Series sum

Given $$\sum_{n=1}^{\infty} \frac{(-1)^n}{2^n}$$ determine ig the series converge or diverge. If convergent calculate the sum.

Using the convergence test I've found out that the series converges. However, I don't know how to go about finding the sum. Any ideas?

• It's geometric. – Matthew Leingang Nov 5 '19 at 2:31
• Note that $\frac{(-1)^n}{2^n} = \left(-\frac{1}{2}\right)^n$. – KM101 Nov 5 '19 at 2:33
• Welcome here! Which convergence test did you use? – Taladris Nov 5 '19 at 2:48
• @Taladris the alternant series test – Max Nov 5 '19 at 2:49

Recall: the geometric series $$\sum_{n=0}^\infty x^n$$ is absolutely convergent with value

$$\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$

whenever $$|x| < 1$$. To find the value for your series, notice that you can "pull out" the general series' first term like this:

$$\sum_{n=0}^\infty x^n = 1 + \sum_{n=1}^\infty x^n$$

Then notice your series has $$x = -1/2$$, apply the formula for the series, and solve for yours to obtain its value.

More explicitly,

$$\sum_{n=1}^\infty x^n = \frac{1}{1-x} - 1\;\;\;\;\; \overset{x=-1/2}{\implies} \;\;\;\;\; \sum_{n=1}^\infty \left( - \frac 1 2 \right)^n = - \frac 1 3$$

• I think it's supposed to be $-\dfrac13$. – Don Thousand Nov 5 '19 at 2:44
• Yeah I made a sign error. My bad; thanks – Eevee Trainer Nov 5 '19 at 2:45
• Oooh, I knew $\sum_{n=1}^{\infty}ar^{n-1} = \frac{a}{1-r}$. But the original had $ar^{n}$, when I encounter a problem like that should I just work with $\sum_{n=0}^{\infty}r^{n} = \frac{1}{1-r}$? I just wasted time playing around with the argument to adapt it to $ar^{n-1}$ haha. – Max Nov 5 '19 at 2:57
• I feel like I was told a similar formula for stuff like that in middle school. I feel like $$\sum_{n=0}^\infty r^n = \frac{1}{1-r}$$ is easier to remember, though, and its manipulations are more intuitive as well. Especially considering it amounts to $$1+r+r^2 + r^3 + \cdots$$ when expanded. Your first formula is almost equivalent to my simpler series: just re-index it (let $m=n-1$ and rewrite your series) and divide both sides by $a$. Yours basically accounts for series whose first term is not $1$ (but that can be accounted for by factoring anyways). – Eevee Trainer Nov 5 '19 at 5:21

Let $$a_n = \frac{(-1)^n}{2^n}=\left(-\frac{1}{2}\right)^n$$

Then $$|a_n| = \frac{1}{2^n}$$ and series $$\sum_{n=1}^\infty |a_n|$$

converge because it is geometric series.

So $$\sum a_n$$ is absolutly converge and this make $$\sum a_n$$ also converge.

And now we use $$\sum_{n=1}^\infty x^n = \frac{x}{1-x}$$

put $$x=-\frac{1}{2}$$ and we get

$$\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty\left(-\frac{1}{2}\right)^n = \frac{-\frac{1}{2}}{1+\frac{1}{2}}=\boxed{-\frac{1}{3}}$$