Most of the Convergence tests require the terms of an infinite series to be positive. At times we may have to deal with series whose terms are not necessarily positive.

In particular, it is important to learn the methods for testing the convergence of Alternating Series, the series whose terms alternate back and forth from positive and negative. In other words, an Alternating Series is a series whose terms are alternatively positive and negative.
Mathematically, an alternating series can be either expressed using Sigma notation or written in expanded form.
$\sum_{n=1}^{\infty }(-1)^{n+1}a_{n}$ = $a_1 - a_2 + a_3 - a_4 + .......... + (-1)n-1 an + ............ ∞
Example:
The series $\sum_{n=1}^{\infty }(-1)^{^{n+1}}$ $\frac{1}{n}$ = 1 - $\frac{1}{2}$ + $\frac{1}{3}$ - $\frac{1}{4}$ +........... is the Alternating Harmonic Series. It is interesting to note that the Alternating Harmonic Series is convergent, even though the General Harmonic Series 1 + $\frac{1}{2}$ + $\frac{1}{3}$ + $\frac{1}{4}$ + ............ is known to be a divergent series.

The alternating series $\sum_{n=1}^{\infty }(-1)^{^{n+1}}$ an = a1 - a2 + a3 - a4 + ................ ∞ ( an > 0) is convergent, if
  1. an + 1  $\leq$ an ; for all n
  2. $\lim_{n\rightarrow \infty }$an = 0

Both the conditions are important to establish the convergence of ∑ an. While the second condition fails to ascertain the divergence of an infinite series, at the same time does not say the series is convergent.

Let us look at an Alternating Series which is Divergent.

Determine whether the alternating series $\sum_{k=1}^{\infty }$ $\frac{(-1)^{k}k}{k+4}$ is convergent or divergent.

Ignoring the sign the general term of the alternating series ak = $\frac{k}{k+4}$

$\lim_{k\rightarrow \infty }$ a = $\lim_{k\rightarrow \infty }$ $\frac{k}{k+4}$ = 1 ≠ 0.

The second condition stated in the Alternating Series Test fails.

Hence the given series is divergent.
Earlier we stated that the alternate Harmonic Series 1 - $\frac{1}{2}$ + $\frac{1}{3}$ - $\frac{1}{4}$ + .......... is convergent even though the general Harmonic Series

1 + $\frac{1}{2}$ + $\frac{1}{3}$ + $\frac{1}{4}$ + ........ is a divergent series.

Let us prove this using Alternating Series Test.

Ignoring the sign, each term in the series is a unit fraction. A unit fraction becomes smaller as the whole value increases. Thus we have the first condition for Alternating Series Test is satisfied.

an+1 = $\frac{1}{n+1}$   and an = $\frac{1}{n}$
$\frac{1}{n+1}$ < $\frac{1}{n}$  ⇒ an+1 $\leq$ an.

Also $\lim_{n\rightarrow \infty }$ an = $\lim_{n\rightarrow \infty }$ $\frac{1}{n}$ = 0
As both the conditions are satisfied, the Alternating Harmonic Series converges.

Alternating Harmonic

In the above graph each term of the alternating series is plotted against the term index. It can be seen, that the term approaches zero from either side of 0 as the value of n increases.

2. Investigate the convergence of $\sum_{n=1}^{\infty }$ $\frac{cos(n\pi )}{\sqrt{n}}$
   This is an alternating series as the value of cos (nπ) alternates between -1 and 1

   The above series can be rewritten as
   $\sum_{n=1}^{\infty }$ $\frac{(-1)^{n}}{\sqrt{n}}$ 

   As the value of denominator increases as the series advances in terms, we have

   an+1  $\leq$ an (The first condition for Alternating Series is satisfied)

   $\lim_{n\rightarrow \infty }$ a = $\lim_{n\rightarrow \infty }$ $\frac{1}{\sqrt{n}}$ = 0
.
   The second condition is also satisfied.

   Hence the alternating infinite series $\sum_{n=1}^{\infty }$ $\frac{cos(n\pi )}{\sqrt{n}}$ is convergent.