Jayleen Aguirre

2022-04-08

I'm trying to determine the convergence of this series:

$\sum _{n=1}^{\infty}(\frac{{\displaystyle 1}}{{\displaystyle 2}}\xb7\frac{{\displaystyle 3}}{{\displaystyle 4}}\xb7\frac{{\displaystyle 5}}{{\displaystyle 6}}\xb7...\frac{{\displaystyle 2n-3}}{{\displaystyle 2n-2}}\xb7\frac{{\displaystyle 2n-1}}{{\displaystyle 2n}}{)}^{a}$

stecchiniror7

Beginner2022-04-09Added 14 answers

Let

${a}_{n}=\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \dots \cdot \frac{2n-3}{2n-2}\cdot$

Note that $a}_{n+1}={a}_{n}\cdot \frac{2n+1}{2(n+1)$

We will show that for all $n$:

$\frac{1}{\sqrt{4n}}\le {a}_{n}\le \frac{1}{\sqrt{2n+1}}$

Having established this, it will follow by comparison with $p$-series that $\sum _{n=1}^{\mathrm{\infty}}{a}_{n}^{a}$

converges if and only if $a>2$

We establish the inequalities by induction.

Towards showing that $a}_{n}\ge \frac{1}{\sqrt{2n+1}$ first note that $a}_{1}\le \frac{1}{\sqrt{2\cdot 1+1}$

Now suppose $a}_{n}\le \frac{1}{\sqrt{2n+1}$. Then

$a}_{n+1}^{2}\le {\left[\frac{2n+1}{2(n+1)}\right]}^{2}\cdot \frac{1}{2n+1}=\frac{2n+1}{4{(n+1)}^{2}}=\frac{2n+1}{4{n}^{2}+8n+4$

$\le \frac{2n+1}{4{n}^{2}+8n+1}=\frac{1}{2n+1}$

So, $a}_{n}\le \frac{1}{\sqrt{2n+1}$ for all n

Towards showing that $a}_{n}\ge \frac{1}{\sqrt{4n}$. Then $a}_{n+1}^{2}\ge {\left[\frac{2n+1}{2(n+1)}\right]}^{2$$\xb7\frac{1}{4n}$. Now

${\left[\frac{2n+1}{2(n+1)}\right]}^{2}\cdot \frac{1}{4n}-\frac{1}{4(n+1)}=\frac{1}{16n{(n+1)}^{2}}>0$

thus

$a}_{n+1}^{2}\ge \frac{1}{4(n+1)$

So, $a}_{n}\ge \frac{1}{\sqrt{4n}$ for all n.

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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