# Series Convergence Alternating Series

The sum of a series is defined as the limit of the sequence of partial sums, $$\{S_N\}$$, which means

$$1-\frac1 2+ \frac 1 4-\frac 1 8+\frac 1 {16}-\cdots =\lim_{n\to \infty} \text{ ?}$$

I got that the sum is $$\sum_{n=0}^{\infty} (-1)^n{\frac1 {2^n}}$$, but I dont know how to find the limit of the sequence of partial sums.

I tried $$\lim_{n\to \infty}$$ $$\frac {1+(1/2)^n} {1-(1/2)}$$ =2 but it did not work.

• Hint: Geometric series. – John Omielan Mar 8 at 22:28
• I tried $$\lim_{x \ to infty} {\frac {1+(\frac 1 2)^n} {1-(1/2)}=2$$ but it did not work. – Tony Mar 8 at 22:41
• You have the right idea, but you have a small mistake in the numerator and, more importantly, you're using $r = \frac{1}{2}$ instead of $r = -\frac{1}{2}$ in the denominator. The correct limit of the partial sums is $\lim_{n \to \infty}\frac{1-\left(-\frac{1}{2}\right)^n}{1-\left(-\frac{1}{2}\right)} = \frac{1}{\left(\frac{3}{2}\right)} = \frac{2}{3}$. – John Omielan Mar 8 at 22:48
• Why does r=-1/2? – Tony Mar 8 at 22:53
• It's $r = -\frac{1}{2}$ because that is what you multiply each term by to get the next term. For example, $-\frac{1}{2} = -\frac{1}{2}(1)$, $\frac{1}{4} = -\frac{1}{2}\left(-\frac{1}{2}\right)$, $-\frac{1}{8} = -\frac{1}{2}\left(\frac{1}{4}\right)$, etc. You can also see this from your general term of $t_n = (-1)^n\frac{1}{2^n} = \left(-\frac{1}{2}\right)^n$, so $\frac{t_{n+1}}{t_n} = -\frac{1}{2}$. Also, note your question currently has one close vote as "Needs details or clarity". I suggest you add what you tried in your earlier comment to your question to help avoid it being closed. – John Omielan Mar 8 at 22:58

Let $$x\in\mathbb R$$ with $$|z|<1$$ and define $$S_n = \sum_{k=1}^n x^k$$. Then $$S_n -xS_n = \sum_{k=1}^n x^k - \sum_{k=1}^n x^{k+1} = 1 - x^{n+1}.$$ Dividing by $$1-x$$, we have $$S_n = \frac{1-x^{n+1}}{1-x}.$$ Choose a positive integer $$N>\frac{\log \varepsilon}{\log|x|}$$, then for $$n\geqslant N$$ we have $$x^{n+1}<\varepsilon$$, and hence $$\lim_{n\to\infty} S_n = \frac1{1-x}$$.
Here $$x=-\frac12$$, so $$\sum_{n=0}^\infty \left(-\frac12\right)^n = \frac1{1+\frac12} = \frac23.$$ 