Calculus of a Single Variable, Chapter 9, 9.3, Section 9.3, Problem 77

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.