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Evaluate if the following series is convergent or divergent: $\sum\limits_n^\infty\frac 1 {\sqrt{n(n+1)}}$.

I could use the integral test that would prove me this series to be divergent. However I want to prove them divergent using Weierstrass comparison theorem.

$$\sum_n^\infty \frac{1}{\sqrt{n(n+1)}}>\sum_n^\infty \frac 1 {n(n+1)} = \sum_n^\infty \frac{1}{n^2+n}=\text{?}$$

However I cannot find a series that are smaller than the last. I tried to find any inequality to bring $n^2$ down to $n$, but I was not successful.

Question:

How can I find a smaller divergent series for $\sum_\limits{n}^{\infty}\frac{1}{{n^2+n}}$?

Thanks in advance!

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  • $\begingroup$ $\displaystyle{1 \over \,\sqrt{\,n\left(n + 1\right)\,}\,} \sim {1 \over n}\ \mbox{as}\ n \to \infty.$ $\endgroup$ – Felix Marin Mar 31 '18 at 5:42
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Since $n(n+1)<4n^2$ for every natural $n$, you have$$(\forall n\in\mathbb{N}):\frac1{\sqrt{n(n+1)}}>\frac1{2n}.$$

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  • $\begingroup$ Thanks for your answer! How do you prove the inequality $n(n+1)<4n^2$? I have already tried to use the binomial but I did not succeed. $\endgroup$ – Pedro Gomes Mar 30 '18 at 16:55
  • $\begingroup$ @PedroGomes $$n(n+1)<4n^2\iff 3n^2-n>0\iff n(3n-1)>0,$$which is true, since $n,3n-1>0$. $\endgroup$ – José Carlos Santos Mar 30 '18 at 16:57
  • $\begingroup$ I apologise for having expressed myself wrong. Is there no intuitive way to derive the inequality? $\endgroup$ – Pedro Gomes Mar 30 '18 at 17:01
  • $\begingroup$ @PedroGomes You can say that the inequality is trivial if $n=1$ and that clearly $4n^2$ growths faster than $n(n+1)$. $\endgroup$ – José Carlos Santos Mar 30 '18 at 17:05
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HINT: $$ \sum_{n=1}^\infty \frac1{\sqrt{n(n+1)}}>\sum_{n=1}^\infty \frac1{\sqrt{(n+1)^2}} $$

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Showing that the $n$th term exceeds $\dfrac 1 {n(n+1)}$ will not show the series diverges, because the series whose terms are $\dfrac 1 {n(n+1)}$ converges. $$ \frac 1 {\sqrt{n(n+1)}} > \frac 1 {\sqrt{(n+1)^2}} = \frac 1 {n+1}, \text{ and } \sum_{n=1}^\infty \frac 1 {n+1} = +\infty. $$

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(1). $\sum \frac {1}{n^2+n}$ converges. But $n^2+n<(n+1)^2$ so $\frac {1}{\sqrt {n^2+n}}\;>\frac {1}{n+1}>0$. And $\sum \frac {1}{n+1}$ diverges.

(2). Suppose $a_n>0$ and $c=\lim_{n\to \infty}\inf_{m>n} (b_m/a_m)>0. $ If $\sum a_n$ diverges then $\sum_n b_n$ diverges. Because for all but finitely many $n$ we have $2b_n/c>a_n,$ so $\frac {2}{c}\sum b_n$ diverges..... In your Q let $a_n=1/n$ and $b_n=1/\sqrt {n^2+n}\;$. Then $b_n/a_n=1/\sqrt {1+1/n}$ which converges to $1$. So, a fortiori, $\lim \sup\; b_n/a_n=1.$

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Of course $\frac{1}{\sqrt{n(n+1)}}>\frac{1}{n+1}$, so the given series is divergent by comparison with the harmonic series.
A fancy approach is to employ the Cauchy-Schwarz inequality $$ \log\left(\frac{n+1}{n}\right) = \int_n^{n+1}\frac{dx}{x}\stackrel{\text{CS}}{\leq} \sqrt{\int_{n}^{n+1}\frac{dx}{x^2}}=\frac{1}{\sqrt{n(n+1)}} $$ such that the explicit lower bound $$ \sum_{n=1}^{N}\frac{1}{\sqrt{n(n+1)}} \geq \log(N+1) $$ follows by telescoping.

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As an alternative note that

$$\frac{1}{\sqrt{n(n+1)}}\sim \frac 1n$$

thus the given series diverges by limit comparison test with $\sum \frac 1n$.

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