Does the Series \(\sum_{n=1}^{\infty} \left[n(n+1)\right]^{-1/2}\) Converge?

[ognik]

Use the comparison test to see if \(\displaystyle \sum_{1}^{\infty}{\left[n\left(n+1\right)\right]}^{-\frac{1}{2}} \)converges?

I tried \(\displaystyle n+1 \gt n, \therefore n(n+1) \gt n^2 , \therefore {\left[n(n+1)\right]}^{\frac{1}{2}} \gt n, \therefore {\left[n(n+1)\right]}^{-\frac{1}{2}} \lt \frac{1}{n}\) - no conclusion possible here

 

[Guest2]

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

 

[Opalg]

Use the comparison test to see if \(\displaystyle \sum_{1}^{\infty}{\left[n\left(n+1\right)\right]}^{-\frac{1}{2}} \)converges?

I tried \(\displaystyle n+1 \gt n, \therefore n(n+1) \gt n^2 , \therefore {\left[n(n+1)\right]}^{\frac{1}{2}} \gt n, \therefore {\left[n(n+1)\right]}^{-\frac{1}{2}} \lt \frac{1}{n}\) - no conclusion possible here

Instead of $n+1 > n$ as a starting point, try it in the form $n < n+1$. That will give you $[n(n+1)] < (n+1)^2.$ Where does that lead you?

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

Not true: $\sqrt{n(n+1)} \le n(n+1)$, but when you take the reciprocals the inequality goes the other way round.

 

[ognik]

So the final comparison is with \(\displaystyle \frac{1}{n+1}\) - interesting the difference that small change in approach makes - the take away for me is that when I get an inconclusive inequality, I immediately try approaching it w.r.t. the 'other direction' inequality :-)

When justifying my answer, is it enough to refer to the p series with p = 1? Or do I need to be more rigorous - my understanding is that with comparison tests, \(\displaystyle \frac{1}{n+1}\) is the same thing as \(\displaystyle \frac{1}{n}\) ?

 

[Opalg]

So the final comparison is with \(\displaystyle \frac{1}{n+1}\) - interesting the difference that small change in approach makes - the take away for me is that when I get an inconclusive inequality, I immediately try approaching it w.r.t. the 'other direction' inequality :-)

Yes, it's always good to look at such problems from both sides. (Nod)

When justifying my answer, is it enough to refer to the p series with p = 1? Or do I need to be more rigorous - my understanding is that with comparison tests, \(\displaystyle \frac{1}{n+1}\) is the same thing as \(\displaystyle \frac{1}{n}\) ?

It should be enough to say that this is effectively the same as $\sum 1/n$. The only difference is that the first term is missing. But convergence is unaffected by changing a finite number of terms at the start of a series.

 

[Prove It]

Use the comparison test to see if \(\displaystyle \sum_{1}^{\infty}{\left[n\left(n+1\right)\right]}^{-\frac{1}{2}} \)converges?

I tried \(\displaystyle n+1 \gt n, \therefore n(n+1) \gt n^2 , \therefore {\left[n(n+1)\right]}^{\frac{1}{2}} \gt n, \therefore {\left[n(n+1)\right]}^{-\frac{1}{2}} \lt \frac{1}{n}\) - no conclusion possible here

For $\displaystyle \begin{align*} n \geq 2 \end{align*}$ we have $\displaystyle \begin{align*} n^2 + n < 2\,n^2 \end{align*}$, so

$\displaystyle \begin{align*} n^2 + n &< 2\,n^2 \\ \sqrt{ n^2 + n } &< \sqrt{ 2\,n^2 } \\ \sqrt{ n \left( n + 1 \right) } &< \sqrt{2}\,n \\ \frac{1}{\sqrt{ n \left( n + 1 \right) } } &> \frac{1}{\sqrt{2}\,n} \end{align*}$

thus

$\displaystyle \begin{align*} \sum_{n = 1}^{\infty} \frac{1}{\sqrt{ n \left( n + 1 \right) }} &> \sum_{n = 1}^{\infty} \frac{ 1}{\sqrt{2}\,n} \\ &= \frac{1}{\sqrt{2}} \sum_{ n = 1} ^{\infty} \frac{1}{n} \end{align*}$

Since $\displaystyle \begin{align*} \sum_{n = 1}^{\infty} \frac{1}{n} \end{align*}$ is divergent, so is $\displaystyle \begin{align*} \frac{1}{\sqrt{2}}\sum_{n = 1}^{\infty} \frac{1}{n} \end{align*}$ and thus $\displaystyle \begin{align*} \sum_{ n = 1} ^{\infty} \frac{1}{\sqrt{n \left( n + 1 \right) } } \end{align*}$ is divergent by comparison.