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

To verify if this infinite series diverges, apply the Divergent Test.

In the Divergence Test, it states that if the limit of `a_n` is not zero, or does not exist, then the sum diverges.

`lim_(n->oo) a_n != 0` or `lim_(n->oo) = DNE`

`:.` `sum ` `a_n` diverges

So, taking the limit of `a_n` as n approaches infinity yields:

`lim_(n->oo) a_n`

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

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

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

`=1/(1+0)`

`=1`

Since the result is not equal to zero, **therefore, the series is divergent**.