`sum_(n=1)^oo n/sqrt(n^2+1)` Verify that the infinite series diverges

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`sum_(n=1)^oo n/sqrt(n^2+1)`

To verify if the series diverges, apply the nth-Term Test for Divergence.

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) a_n =DNE`

`:.` `sum` `a_n`  diverges

Applying this, the limit of...

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`sum_(n=1)^oo n/sqrt(n^2+1)`

To verify if the series diverges, apply the nth-Term Test for Divergence.

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) a_n =DNE`

`:.` `sum` `a_n`  diverges

Applying this, the limit of the term of the series as n approaches infinity is:

`lim_(n->oo) a_n`

`=lim_(n->oo) n/sqrt(n^2+1)`

`=lim_(n->oo) n/sqrt(n^2(1+1/n^2))`

`=lim_(n->oo) n/(nsqrt(1+1/n^2))`

`=lim_(n->oo) 1/sqrt(1+1/n^2)`

`=(lim_(n->oo)1)/(lim_(n->oo)sqrt(1+1/n^2))`

`=1/sqrt(0+1)`

`=1`

The limit of the series is not zero. Therefore, by the nth-Term Test forDivergence, the series diverges.

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