# What is the limit for this series

What is the limit: $$\sum_{n=1}^{\infty}\left(\frac{1}{2}\right)^{n-1}$$

Can I just say its $1^\infty / 2^\infty$ and it goes to zero? That doesn't seem right? This question springs from me wondering why I cannot do the limit test to see if this series converges or diverges.

I know that I can easily find that this is a geometric series that converges to $2$. But I was wondering why we cannot do the nth term test for divergence.

Thanks

-
The terms do have limit $0$. (But you shouldn't "plug in" $\infty$.) However, knowing the terms approach $0$ does not imply convergence. The standard example is $\sum_1^\infty\frac{1}{n}$, which diverges. –  André Nicolas Aug 9 '12 at 4:36

Theorem:

If the series $\sum_{k=1}^{\infty}a_k$ converges, then

$$\displaystyle\lim_{n\to\infty}a_n=0$$

Proof:

From the convergence of the sum, we know that the sequence of partial sums $\{S_n\}$ converges. Say $\displaystyle\lim_{n\to\infty}S_n=S$. From $a_n=S_n-S_{n-1}$, we will get: $$\displaystyle\lim_{n\to\infty}a_n=\displaystyle\lim_{n\to\infty}(S_n-S_{n-1})=S-S=0$$ Q.E.D

Now, is the opposite direction -- If $\displaystyle\lim_{n\to\infty}a_n=0$, then $\sum_{k=1}^{\infty}a_k$ converges, will work too? The answer is no!

Take the harmonic series, $\sum_{n=1}^{\infty}\frac{1}{n}$, as you can see $\displaystyle\lim_{n\to\infty}\frac{1}{n}=0$, but is the $\sum_{n=1}^{\infty}\frac{1}{n}$ converges? No! And why not? Here:

First, Cauchy criteria:

$\sum_{k=1}^{\infty}a_k$ converges iff

$$\forall\epsilon>0, \exists n, n>N(\epsilon), \forall p\in\mathbb{N}$$ $$|a_{n+1}+a_{n+2}+...+a_{n+p}|=|\sum_{k=n+1}^{n+p}a_k|<\epsilon$$

Applying above on $\sum_{k=1}^{\infty}\frac{1}{k}$:

Take $p=n$,

$$S_{2n}-S_n=\sum_{k=n+1}^{2n}\frac{1}{k}=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}>n \frac{1}{2n}=\frac{1}{2}$$

Hence, $S_{2n}-S_n$ can't be equal less than given $\epsilon$. And therefor the series is diverges.

-

A geometric series converges when $|r| < 1$, in which case

$$\sum_{k=0}^{\infty}ar^k - r\sum_{k=0}^{\infty}ar^k = a$$

so

$$\sum_{k=0}^{\infty}ar^k = \frac{a}{1-r}$$

The restriction on $r$ is required since

$$\sum_{k=0}^{\infty}ar^k = \lim_{n \to \infty} \sum_{k=0}^{n}ar^k = \lim_{n \to \infty}\frac{a(1 - r^{n+1})}{1-r} = \frac{a}{1-r}$$

only when $|r| < 1.$ This formula tells you that your geometric series converges to $\frac{1}{1 - 1/2} = 2$

You cannot conclude that a series converges by using the n'th term test.

-

According to this, the sum is $\frac{1}{1-\frac{1}{2}}=2$ as the first term is 1 and the common ration is $\frac{1}{2}$

-