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

MATH101 April 2014

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Question 07 (a)
Long Problem. Show your work. No credit will be given for the answer without the correct accompanying work. Determine, with explanation, whether the following series converges or diverges. $\sum _{n=1}^{\infty }{\frac {1}{\sqrt {n^{2}+1}}}$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
We know a lot of convergence information about series of the form ${\frac {1}{n^{p}}}$. Can we use this as a comparison in anyway to ${\frac {1}{\sqrt {n^{2}+1}}}$?

Hint 2
Notice if we take $n$ to be large then ${\frac {1}{\sqrt {n^{2}+1}}}\approx {\frac {1}{\sqrt {n^{2}}}}={\frac {1}{n}}.$ What is the convergence behaviour of ${\frac {1}{n}}$? This should give us a hint of what kind of series we should try and compare to.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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.

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