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Prove that the series is convergent:

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

Ratio test and square-root theorem seem useless here, so I have tried using direct comparison test.

$$\frac{\sqrt{n^{2}+1}}{n^{2}}< \frac{n^{2}+1}{n^{2}} = \frac{n^{2}}{n^{2}}+\frac{1}{n^{2}}= 1+\frac{1}{n^{2}}$$

$$\Rightarrow 1+\sum_{n=1}^{\infty}\frac{1}{n^{2}}$$

And we know this kind of series will converge, so the original one will converge as well.


It was hard for me to exclude ratio test & co. before testing them on paper, so it was very time consuming for sure. Cannot afford that in the exam...

Did I do it correctly at least?

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    $\begingroup$ This is divergent $\endgroup$ – iamvegan Sep 19 '16 at 16:59
  • $\begingroup$ No, $\sum_{n=1}^\infty (1 + \frac{1}{n^2}) \neq 1 + \sum_{n=1}^\infty \frac{1}{n^2}$. $\endgroup$ – Jair Taylor Sep 19 '16 at 16:59
  • $\begingroup$ You made a big mistake in your derivation: $1$ is under the summation sign, so it becomes $\infty$. Then your comparison test is inconclusive. $\endgroup$ – Yves Daoust Sep 19 '16 at 17:00
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    $\begingroup$ $$\frac{\sqrt{n^2 + 1}}{n^2} \sim \frac{1}{n}$$ $\endgroup$ – ÍgjøgnumMeg Sep 19 '16 at 17:00
  • $\begingroup$ Series is not convergent. The square root in the numerator is larger than n, so the series is bounded below by a harmonic series. $\endgroup$ – marioga Sep 19 '16 at 17:01
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This series does not converge because $\sqrt{n^2+1}>n$ and then $$ \frac{\sqrt{n^2+1}}{n^2}>\frac{n}{n^2}=\frac{1}{n} $$ and we know that $$ \sum_{n\geq 1} \frac{1}{n}=+\infty $$

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  • $\begingroup$ Ouch absolutely right, I had a thinking mistake and thought the square root would erase it entirely and leave 1 in numerator.. :D Thank you and everyone else too, understood! $\endgroup$ – cnmesr Sep 19 '16 at 17:04
  • $\begingroup$ Sometimes, it helps to have an external eye <) $\endgroup$ – Duchamp Gérard H. E. Sep 19 '16 at 17:05
  • $\begingroup$ Sure, I was just afk. By the way, you got any profit if I accept your answer? There is some kind of bonus credit you get? I always wanted know that ^^ $\endgroup$ – cnmesr Sep 19 '16 at 17:45
  • $\begingroup$ @cnmesr [By the way, you got any profit if I accept your answer?]--->yes, I get reputation points. In fact, I do practice principally for fun and to help, but if fun can reward, why not ? $\endgroup$ – Duchamp Gérard H. E. Sep 19 '16 at 18:46
  • $\begingroup$ How many you get? I've never answered a question, all my points you see come from questions so no idea :D $\endgroup$ – cnmesr Sep 19 '16 at 19:53
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The series you wrote is not convergent: \begin{align*} \frac{\sqrt{n^2+1}}{n^2} \ge \frac{\sqrt{n^2}}{n^2} = \frac{n}{n^2} = \frac{1}{n}, \end{align*}

and $\sum_{n \ge 1} \frac1n$ diverges.


Your reasoning goes wrong when you go from $$\sum_{n \ge 1} \left( 1+\frac{1}{n^{2}} \right)$$

to $$ 1 + \sum_{n \ge 1} \frac{1}{n^2}. $$

The problem of course is that you are pulling out a $1$ for each $n$, so the first term is actually $\sum_{n \ge 1} 1 = 1 + 1 + 1 + \cdots = \infty$. So you only showed that your series was bounded above by a divergent series.

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$$ \frac{\sqrt{n^2+1}}{n^2} > \frac{\sqrt{n^2}}{n^2} = \frac{1}{n} $$ And the harmonic series is divergent.

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$$\sum _{ n=1 }^{ \infty } \frac { \sqrt { n^{ 2 }+1 } }{ n^{ 2 } } >\sum _{ n=1 }^{ \infty } \frac { \sqrt { n^{ 2 } } }{ n^{ 2 } } =\sum _{ n=1 }^{ \infty } \frac { 1 }{ n } $$ serie is divergent

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we have $$\frac{n\sqrt{1+1/n^2}}{n^2}=1/n\sqrt{1+1/n^2}>1/n$$

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In such series, you can often get an idea of the asymptotic behavior as a power of $n$.

In this case, when $n$ grows, $\sqrt{n^2+1}\approx\sqrt{n^2}=n$ and $\sqrt{n^2+1}/n^2\approx1/n$. This latter series is known to diverge.


More generally,

$$\frac{(P(n))^a}{(Q(n))^b}$$ where $P$ and $Q$ are polynomials in $n$ of degree $p$ and $q$ canbe approximated by their leading terms

$$\frac{(P_pn^p)^a}{(Q_qn^q)^b}=Cn^{pa-qb}$$ which give you a hint on convergence.

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$$\frac{\sqrt{n^2+1}}{n^2} = \sqrt{\frac{n^2+1}{n^4}} = \sqrt{\frac{1}{n^2} + \frac{1}{n^4}} = \frac{1}{n}\sqrt{1 + \frac{1}{n^2}}$$

That is by comparison divergent because of the harmonic series.

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You asked if you did this correctly.

You were right up to where you proved that, for all positive integers $n$, we have $$\frac{\sqrt{n^2+1}}{n^2} < 1+\frac{1}{n^2}$$

You made one mistake when you looked at the sum. We should have \begin{eqnarray*}\sum_{n=1}^{\infty} \frac{\sqrt{n^2+1}}{n^2} \ \ &<& \ \ \ \sum_{n=1}^{\infty} \left(1 + \frac{1}{n^2}\right) \\ \\ &=& \left(\sum_{n=1}^{\infty} 1 \right) + \left(\sum_{n=1}^{\infty}\frac{1}{n^2} \right) \end{eqnarray*}

The problem is that the sum $1+1+1+1+\cdots$ diverges and so the series on the right diverges.

Just because the given series is term-by-term smaller than a divergent series, it does not mean that the given series converges. For example: $2<3$ but both the series $2+2+2+2+\cdots$ and the series $3+3+3+3+\cdots$ diverge. It is false to say that $2<3$, and $3+3+3+3+\cdots = \infty$ and hence $2+2+2+2+\cdots < \infty$, meaning that $2+2+2+2+\cdots$ converges.

The comparison test only works if you can show that one series is smaller than a convergent series.

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