Science:Math Exam Resources/Courses/MATH101/April 2014/Question 07 (a)/Solution 1

Note that ${\frac {1}{\sqrt {n^{2}+1}}}$ is similar to ${\frac {1}{\sqrt {n^{2}}}}={\frac {1}{n}}$ for large $n$. The series of $1/n$ diverges and so, we expect that the series we are interested in diverges as well. It will be helpful if we can use the comparison test to compare to the $1/n$ series. To do this we need to find a sequence related to $1/n$ that is greater than the sequence we have.

Notice that for $n\geq 1$,

${\begin{aligned}{\sqrt {n^{2}+1}}\leq {\sqrt {n^{2}+n^{2}}}={\sqrt {2}}n\end{aligned}}.$

If we invert this then the inequality changes,

${\begin{aligned}{\frac {1}{\sqrt {n^{2}+1}}}\geq {\frac {1}{{\sqrt {2}}n}}\end{aligned}}.$

The left of this expression is the set of terms we wish to sum and on the right is a sequence involving $1/n$. The terms we wish to sum are larger than the terms that we know diverge. This is precisely what we wanted and therefore, if we compare the series then

${\begin{aligned}\sum _{n=1}^{\infty }{\frac {1}{\sqrt {n^{2}+1}}}\geq \sum _{n=1}^{\infty }{\frac {1}{{\sqrt {2}}n}}={\frac {1}{\sqrt {2}}}\sum _{n=1}^{\infty }{\frac {1}{n}}\end{aligned}}$

Since the series $\sum _{n=1}^{\infty }{\frac {1}{n}}$ diverges then, by the comparison test, $\sum _{n=1}^{\infty }{\frac {1}{\sqrt {n^{2}+1}}}$ diverges as well.