Test the convergence and absolute convergence of the following series:

Test the convergence and absolute convergence of the alternating harmonic series.

1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...

Test the convergence and absolute convergence of the series:

1-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+...

Test the convergence and absolute convergence of the series:

\sum_{n=2}^{\infty}(-1)^{n+1}\frac{1}{log\,n}

Solution: (i) The given series

\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n}

is convergent by Leibnitz’s test.

Now, \sum_{n=1}^{\infty}\left | u_n \right |=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...=\sum_{n=1}^{\infty}\frac{1}{n}

is not convergent (As p=1)

Hence the given series is not absolutely convergent. This is an example of conditionally convergent series.

(ii) The given series

\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n^2}

is convergent by Leibnitz’s test.

Also, \sum_{n=1}^{\infty}\left | u_n \right |=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...=\sum_{n=1}^{\infty}\frac{1}{n^2}

is convergent (As p = 2 > 1) Hence the given series is absolutely convergent.

The given series

\sum_{n=2}^{\infty}(-1)^{n+1}U_n

Here u_n=\frac{1}{log\,n}

Now log x is an increasing function

\forall\; x>0

:log(n+2)>log(n+1)

or \;\;\frac{1} {log(n+2)}<\frac{1} {log(n+1)}

\therefore u_{n+1} \leq u_n

Also \lim_{n\rightarrow \infty}u_n=\lim_{n\rightarrow \infty}\frac{1}{log\,n}=0

Hence by Leibnitz’s test , the given series is convergent.
Now for absolute convergence , consider

\sum_{n=1}^{\infty}\left  | u_n \right | = \sum_{n=1}^{ \infty} \frac{1} {log\,n}

It is a divergent series Hence the given series is not absolutely convergent. This is an example
of conditionally convergent series.

keywords

absolute convergence
determine whether the series is absolutely convergent, conditionally convergent, or divergent.
absolute convergence test
every absolutely convergent series is
absolutely convergent series examples
absolute convergence vs conditional convergence