Calculus of a Single Variable, Chapter 9, 9.4, Section 9.4, Problem 19

Recall Limit Comparison Test considers two positive series a_ngt=0 and b_ngt 0 for all n such that the limit from the ratio of two series as:
lim_(n-gtoo)a_n/b_n=c
where c is positive and finite ( 0ltcltoo) .
When we satisfy the condition for the limit value, the two series will have the same properties. Both will either converge or diverges. We may also consider the conditions:
If we have lim_(n-gtoo)a_n/b_n=0 , we follow: sum b_n converges then sum a_n converges.
If we have lim_(n-gtoo)a_n/b_n=oo , we follow: sum b_n diverges then sum a_n diverges.
For the given series sum_ (n=1)^oo 1/(nsqrt(n^2+1)) , we may let a_n= 1/(nsqrt(n^2+1)).
Rationalize the denominator:
1/(nsqrt(n^2+1)) *sqrt(n^2+1)/sqrt(n^2+1) =sqrt(n^2+1)/(n(n^2+1)
Note: sqrt(n^2+1)*sqrt(n^2+1) = (sqrt(n^2+1))^2 = n^2+1 .
Ignoring the constants, we get:
sqrt(n^2+1)/(n(n^2+1)) ~~sqrt(n^2)/(n(n^2)) or 1/n^2
Note: sqrt(n^2) =n . We may cancel it out to simplify.
This gives us a hint that we may apply comparison between the two series: a_n= 1/(nsqrt(n^2+1)) and b_n = 1/n^2 .
The limit of the ratio of the two series will be:
lim_(n-gtoo) [1/(nsqrt(n^2+1))]/[1/n^2] =lim_(n-gtoo) 1/(nsqrt(n^2+1))*n^2/1
=lim_(n-gtoo) n^2/(nsqrt(n^2+1))
=lim_(n-gtoo) n/sqrt(n^2+1)
Apply algebraic techniques to evaluate the limit. We divide by n with the highest exponent which is n or n^1 . Note: n is the same as sqrt(n^2) .
lim_(n-gtoo) n/(sqrt(n^2+1)) =lim_(n-gtoo) (n/n)/(sqrt(n^2+1)/sqrt(n^2))
=lim_(n-gtoo) 1/sqrt(1+1/n^2)
=1/sqrt(1+1/oo)
= 1/sqrt(1+0)
= 1 /sqrt(1)
= 1/1
=1
The limit value c=1 satisfies 0ltclt oo .
Apply the p-series test: sum_(n=1)^oo 1/n^p is convergent if pgt1 and divergent if plt=1 .
The sum_(n=1)^oo 1/n^2 has p =2 which satisfy pgt1 since 2gt1 . Then, the series sum_(n=1)^oo 1/n^2 is convergent.
Conclusion based from limit comparison test:
With the series sum_(n=1)^oo 1/n^2 convergent, it follows the series sum_ (n=1)^oo 1/(nsqrt(n^2+1)) is also convergent.