`sum_(n=1)^oo ((-1)^(n+1))/(n+1)`

Take note that an alternating series

`sum` `a_n = sum (-1)^(n+1) b_n`

is convergent if the following conditions are satisfied.

(i) `b_n` is decreasing, and

(ii) ` lim_(n->oo) b_n=0` .

In the given alternating series, the bn is:

`b_n = 1/(n+1)`

Then, check if the values of bn...

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`sum_(n=1)^oo ((-1)^(n+1))/(n+1)`

Take note that an alternating series

`sum` `a_n = sum (-1)^(n+1) b_n`

is convergent if the following conditions are satisfied.

(i) `b_n` is decreasing, and

(ii) ` lim_(n->oo) b_n=0` .

In the given alternating series, the bn is:

`b_n = 1/(n+1)`

Then, check if the values of bn decrease as n increases by 1.

`n=1` , `b_n = 1/2`

`n=2` , `b_n=1/3`

`n=3` , `b_n=1/4`

`n=4` , `b_n=1/5`

So bn is decreasing.

Also, take the limit of bn as n approaches infinity.

`lim_(n->oo) b_n = lim_(n->oo) 1/(n+1) = 0`

Since the result is zero, the second condition is satisfied too.

**Therefore, by Alternating Series Test, the given series is convergent.**