`sum_(n=1)^oo n/(n+1)` Verify that the infinite series diverges

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`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...

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`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.

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