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

1 Answer
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 nth partial sum to be the quantity S_{n}=sum_{k=1}^{n}alpha_{k}. This generates a sequence S_{1}, S_{2}, S_{3},..., which we can write in shorthand form as (S_{n})_{n=1}^{infty}.

By definition, the series sum_{k=1}^{infty}alpha_{k} converges if and only if the sequence (S_{n})_{n=1}^{infty} converges.

If lim_{n->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->infty}S_{n}=L.

The series sum_{k=1}^{infty}alpha_{k} diverges if and only if the sequence (S_{n})_{n=1}^{infty} diverges.

Consider, for example, the series sum_{k=1}^{infty}1/(2^k)=1/2+1/4+1/8+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 nth partial sum is S_{n}=sum_{k=1}^{n}1/(2^k)=1/2+1/4+1/8+1/16+ cdots+1/(2^n)=(1/2 (1-(1/2)^(n)))/(1-1/2)=1-1/(2^n).

Clearly, lim_{n->infty}S_{n}=1 for this example, so the infinite series converges and we write sum_{k=1}^{infty}1/(2^k)=1/2+1/4+1/8+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 nth partial sum is S_{n}=sum_{k=1}^{n}2^k=2+4+8+16+cdots+2^n=(2(1-2^n))/(1-2)=2^{n+1}-2.

Clearly lim_{n->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->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}(-1)^{k+1} is a divergent series whose partial sums do not grow without bound.