# Test the convergence and absolute convergence of the following series:

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## 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.

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(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.

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what is absolute convergence of the series

A series Σ a_n converges absolutely if the series of the absolute values, Σ |a_n| converges

what is conditionally convergent series

A series that converges, but does not converge absolutely, converges conditionally.

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