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`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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