# `sum_(n=1)^oo n/sqrt(n^2+1)` Determine the convergence or divergence of the series.

Recall that the Divergence test follows the condition:

If `lim_(n-gtoo)a_n!=0` then `sum a_n` diverges.

For the given series `sum_(n=1)^oo n/sqrt(n^2+1)` , we have `a_n=n/sqrt(n^2+1)`

To evaluate the `a_n=n/sqrt(n^2+1)` , we divide by `n ` with the highest exponent which is `n`  or `sqrt(n^2)` . Note: `n = sqrt(n^2)` .

`a_n=(n/n)/(sqrt(n^2+1)/sqrt(n^2))`

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

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

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

Applying the divergence test, we determine the limit of the series as:

`lim_(n-gtoo)a_n =lim_(n-gtoo)n/sqrt(n^2+1)`

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

` =[lim_(n-gtoo)1] /[lim_(n-gtoo)sqrt(1+1/n^2)]`

` = 1 / sqrt(1+ 1/oo)`

` =1 / sqrt(1+0)`

` =1 / sqrt(1)`

` = 1/1`

` =1`

The `lim_(n-gtoo)n/sqrt(n^2+1)=1` satisfy the condition `lim_(n-gtoo)a_n!=0`.

Therefore, the series `sum_(n=1)^oon/sqrt(n^2+1) ` is a divergent series.

We can also verify with the graph of `f(n) =n/sqrt(n^2+1)` :

As the "n" value increases, the graph diverges.

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