2
$\begingroup$

How does one test for convergence of $\sum_{n=1}^{\infty}\frac{n}{\sqrt{n^2 +1}}$?

My work ...

First I had a look at the series.

$S_n = \frac{1}{\sqrt{2}}+\frac{2}{\sqrt{5}}+\frac{3}{\sqrt{10}}+\frac{4}{\sqrt{17}}+...$

I used Geogebra and the series clearly diverges.

I was going to compare this series to:

$S^{'}_n = 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...$

... but I couldn't see anyway of showing that:

$S_n \ge S^{'}_n$

When I use the ratio test I get:

$\frac{(n+1)\sqrt{n^2+1}}{(n)\sqrt{n^2+2n+2}}$

Which I can't see any way of simplifying.

What method should I use here?

  • 1
    Are you testing convergence of the sequence or of the series ? – 2017-01-12
  • 0
    Convergence of the series. – 2017-01-12
  • 1
    is this a series or the limit of a function - if you divide the numerator and denominator by n, you should be able to see that the expression itself has limit 1 for increasing n - therefore an infinite summation of it has to be unbounded (infinite) – 2017-01-12
  • 1
    the 'ratio test' is used for testing convergence of a summation – 2017-01-12

5 Answers 5

7

$$\frac1{\sqrt{1+\dfrac1{n^2}}}$$ obviously tends to $1$ and the series diverges.

5

Well, the terms don't even go to zero, in fact$$\frac{n}{\sqrt{n^2+1}}\to 1$$ and hence the series doesn't converge.

4

Using the Limit test (Also known as Term Test):

If $\lim_{n \to \infty}a_n\neq 0$, the series $\sum_{n=1}^\infty{a_n}$ diverges.

Thus, let $a_n=\frac{n}{\sqrt{n^2+1}}$.

You will realise that this series has:

$$\lim_{n\to \infty} \frac{n}{\sqrt{n^2+1}}=1$$

Which is not equal to $0$. Therefore, your series diverges.

  • 0
    Hm, this test is known to me by the name [term test](https://en.wikipedia.org/wiki/Term_test), so I don't confuse with the limit comparison test. – 2017-01-12
  • 0
    Hmm, I see. It seems like that name is more common. Perhaps I should change the name on my answer. – 2017-01-12
  • 0
    Well, if Wolfram uses that, it seems it should also be common. And its not like limit comparison test isn't a bad choice here either ;) – 2017-01-12
3

HINT:

$$\frac{n}{\sqrt{n^2+2n+1}}\leq\frac{n}{\sqrt{n^2+1}}\leq\frac{n}{\sqrt{n^2}}$$

3

Recall that if $\lim_{n\to\infty}u_n\neq 0$, then the series $\sum_{n=1}^{\infty}u_n$ diverges.

So, in your question, we have $$\lim_{n\to\infty}\frac{n}{\sqrt{n^2+1}}=\lim_{n\to\infty}\frac{1}{\sqrt{1+\frac{1}{n^2}}}=\frac{1}{\sqrt{1+0}}=1\neq 0.$$

Using the result above, we conclude that the series $\sum_{n=1}^{\infty}\frac{n}{\sqrt{n^2+1}}$ is divergent.