# True or false? A series oo sum_(n=1)alpha_n is convergent if lim_(n->oo)S_n converges. ( S_n = nth partial sum)

Oct 14, 2017

True. This is the definition of what it means for a series to converge.

#### Explanation:

Given an infinite series ${\sum}_{k = 1}^{\infty} {\alpha}_{k}$, we can define the $n$th partial sum to be the quantity ${S}_{n} = {\sum}_{k = 1}^{n} {\alpha}_{k}$. This generates a sequence ${S}_{1} , {S}_{2} , {S}_{3} , \ldots$, which we can write in shorthand form as ${\left({S}_{n}\right)}_{n = 1}^{\infty}$.

By definition, the series ${\sum}_{k = 1}^{\infty} {\alpha}_{k}$ converges if and only if the sequence ${\left({S}_{n}\right)}_{n = 1}^{\infty}$ converges.

If ${\lim}_{n \to \infty} {S}_{n} = L$, we say that ${\sum}_{k = 1}^{\infty} {\alpha}_{k}$ converges to $L$ and write ${\sum}_{k = 1}^{\infty} {\alpha}_{k} = L$. This does not mean we are literally adding up infinitely many terms. It just means ${\lim}_{n \to \infty} {S}_{n} = L$.

The series ${\sum}_{k = 1}^{\infty} {\alpha}_{k}$ diverges if and only if the sequence ${\left({S}_{n}\right)}_{n = 1}^{\infty}$ diverges.

Consider, for example, the series ${\sum}_{k = 1}^{\infty} \frac{1}{{2}^{k}} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots$, which is an (infinite) geometric series.

By the formula for a finite geometric series (see http://www.purplemath.com/modules/series5.htm ), the $n$th partial sum is ${S}_{n} = {\sum}_{k = 1}^{n} \frac{1}{{2}^{k}} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots + \frac{1}{{2}^{n}} = \frac{\frac{1}{2} \left(1 - {\left(\frac{1}{2}\right)}^{n}\right)}{1 - \frac{1}{2}} = 1 - \frac{1}{{2}^{n}}$.

Clearly, ${\lim}_{n \to \infty} {S}_{n} = 1$ for this example, so the infinite series converges and we write ${\sum}_{k = 1}^{\infty} \frac{1}{{2}^{k}} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots = 1$.

As an example of a series that diverges, we consider ${\sum}_{k = 1}^{\infty} {2}^{k} = 2 + 4 + 8 + 16 + 32 + \cdots$. The $n$th partial sum is ${S}_{n} = {\sum}_{k = 1}^{n} {2}^{k} = 2 + 4 + 8 + 16 + \cdots + {2}^{n} = \frac{2 \left(1 - {2}^{n}\right)}{1 - 2} = {2}^{n + 1} - 2$.

Clearly ${\lim}_{n \to \infty} {S}_{n}$ does not exist for this example, so the infinite series ${\sum}_{k = 1}^{\infty} {2}^{k} = 2 + 4 + 8 + 16 + 32 + \cdots$ diverges. Sometimes people might write ${\lim}_{n \to \infty} {S}_{n} = \infty$ and ${\sum}_{k = 1}^{\infty} {2}^{k} = 2 + 4 + 8 + 16 + 32 + \cdots = \infty$ because the partial sums grow without bound for this example.

You should think about why the series ${\sum}_{k = 1}^{\infty} {\left(- 1\right)}^{k + 1}$ is a divergent series whose partial sums do not grow without bound.