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Test the convergence of the series $$1+\frac{2^2}{3^2}+\frac{2^2.4^2}{3^2.5^2}+\frac{2^2.4^2.6^2}{3^2.5^2.7^2}+\cdots$$

I tried solving this problem using Gauss's Test which says,

If $\sum u_n$ is a series of positive real numbers, such that, $\frac{u_n}{u_{n+1}}=1+\frac{a}{n}+\frac{b_n}{n^p},$ where, $p>1$ and $b_n$ is a bounded sequence,then, the series converge if $a>1$, else if $a\leq 1$ the series diverge.

So, the series given here, is $1+\frac{2^2}{3^2}+\frac{2^2.4^2}{3^2.5^2}+\frac{2^2.4^2.6^2}{3^2.5^2.7^2}+\cdots$. Ignoring the first term of the series, we consider $\sum u_n$ where, $u_n=\frac{2^2.4^2.6^2...(2n)^2}{3^2.5^2.7^2.(2n+1)^2},\forall n\in\Bbb N.$

Then we have, $u_{n+1}=\frac{2^2.4^2.6^2...(2n+2)^2}{3^2.5^2.7^2.(2n+3)^2}.$

We note, that $\lim\frac{u_n}{u_{n+1}}=\lim\frac{(2n+3)^2}{(2n+2)^2}=1+\frac 1{n+1}+\frac{\frac 14}{(n+1)^2}.$ From Gauss's Test, we observe, here $p=2\gt 1$ and $b_n=\frac 14$ is a constant and hence a bounded sequence. Also, $a=1,$ due to which the given series is divergent.


I hope that my answer is correct.

However, I don't seem to have an explanation for why I could apply Gauss's Test. This is because, Gauss's Test, say, that $\frac{u_n}{u_{n+1}}$ should be of the following form :$$1+\frac{a}{\color{red}{n}}+\frac{b_n}{\color{red}{n^p}}.$$

But in my solution, $\frac{u_n}{u_{n+1}}$ came out to be of the form :$$1+\frac{a}{\color{green}{n+1}}+\frac{b_n}{\color{green}{(n+1)^p}},$$ (,where $p=2,b_n=\frac 14, a=1$). But then the denominators were of the form $(n+1)$ and not $n$ as was "described" in the test. So, does this create any problem with my solution ? If not, then why it does not matter ?

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  • $\begingroup$ Just study $v_n = u_{n-1}$, it will verify your hypothesis ? $\endgroup$ Commented Jun 15, 2023 at 10:23

5 Answers 5

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Divergence

Taking cross products, we get $$ \left(\frac{2k}{2k+1}\right)^2\ge\frac{2k-1}{2k+1}\tag1 $$ Thus, $$ \frac{2^2}{3^2}\cdot\frac{4^2}{5^2}\cdots\frac{(2n)^2}{(2n+1)^2}\ge\frac1{2n+1}\tag2 $$ Therefore, $$ \begin{align} 1+\frac{2^2}{3^2}+\frac{2^2\cdot4^2}{3^2\cdot5^2}+\frac{2^2\cdot4^2\cdot6^2}{3^2\cdot5^2\cdot7^2}+\cdots &\ge1+\frac13+\frac15+\frac17+\cdots\tag{3a}\\ &\ge\frac12\left(1+\frac12+\frac13+\frac14+\cdots\right)\tag{3b} \end{align} $$ diverges.


Gauss' Test

Let $$ u_0=1,\ u_1=\frac{2^2}{3^2},\ u_2=\frac{2^2\cdot4^2}{3^2\cdot5^2},\ u_3=\frac{2^2\cdot4^2\cdot6^2}{3^2\cdot5^2\cdot7^2}\tag4 $$ then $$ \begin{align} \frac{u_n}{u_{n+1}} &=\left(\frac{2n+1}{2n}\right)^2\tag{5a}\\ &=1+\frac1n+\frac{\frac14}{n^2}\tag{5b} \end{align} $$ So the series diverges.

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It's just shifting of indices.

Define $b_1=1$, and $b_n= u_{n-1}$ for all $n\ge 2$.

Consider the series of positive nos. $\sum_{n=1}^\infty b_n$.

By Gauss' law, for $n\ge 2$, $\frac{b_n}{b_{n+1}}=\frac{u_{n-1}}{u_n}=1+1/n+(1/4)n^{-p}\implies \sum_{n=1}^\infty b_n$ diverges.

This implies that $\sum_{n=1}^\infty b_{n+1}$ diverges. $\sum_{n=1}^\infty b_{n+1}=\sum_{n=1}^\infty u_n$ diverges.

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You can absolutely do the Gauss test by setting $v_n = u_{n-1}$.

If you want to be sure it diverges, I can find you an equivalent. We have

$$w_n = \sqrt{u_n} = \frac{2\cdot 4 \cdots 2n}{1 \cdot 3 \cdots 2n+1} = \frac{2^n n! \cdot 2^n n!}{(2n+1)!} = \frac{2^{2n} (n!)^2}{(2n+1)!}.$$

With Stirling's formula, we get $$w \sim \frac{2^{2n} \left(\sqrt{2 \pi n} \frac{n^n}{e^n}\right)^2}{\sqrt{2 \pi (2n+1)} \frac{(2n+1)^{2n+1}}{e^{2n+1}}} \sim 2^{2n} \frac{2 \pi n}{2\sqrt{\pi n}} \frac{e}{2n+1} \left(\frac{n}{2n+1}\right)^{2n} \sim \frac{e \sqrt{\pi}}{2 \sqrt{n}}\cdot \left( 1 + \frac{1}{2n} \right)^{-2n}$$ Let us study $v_n = \left( 1 + \frac{1}{n} \right)^{-n}$, we have $$v_n = e^{-n\ln\left(1+ \frac{1}{n} \right)} = e^{-n \left(\frac{1}{n} +O\left(\frac{1}{n^2}\right)\right)} = e^{-1 + O \left(\frac{1}{n} \right)}\rightarrow e^{-1}$$ So we get $$w_n \sim \frac{1}{2} \sqrt{\frac{\pi}{n}}$$ Hence $$u_n \sim \frac{\pi}{4n}$$ which is clearly a term of a divergent series. You even have $$S_n \sim \frac{\pi}{4} \ln(n).$$

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I found this connection interesting:

The given series is $_3F_2(1,1,1;\frac32,\frac32;1)$ and by using the integral representation $$_3F_2(a_1,a_2,a_2;b_1,b_2;z)=\frac{\Gamma(b_1)}{\Gamma(a_1)\Gamma(b_1-a_1)}\int_0^1t^{a_1-1}(1-t)^{b_1-a_1-1}\,_2F_1(a_2,a_3;b_2;zt)dt$$ with some conditions on the parameters and $a_1=a_3=a_3=z=1, b_1=b_2=\frac32$ we have

$$_3F_2(1,1,1;\frac32,\frac32;1)=\frac12\int_0^1\frac{\arcsin t}{\sqrt{t}(1-t)}dt$$ which is divergent.

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Here is another proof of diverge using Orrin Frink's test

The $n$-th term of the series is $$a_n=\Big(\frac{(2n)!!}{(2n+1)!!}\Big)^2$$ and so $\Big(\frac{a_{n+1}}{a_n}\Big)^n=\Big(\frac{2n}{2n+1}\Big)^{2n}$

Since $\Big(\frac{2n+1}{2n}\Big)^{2n}=\Big(1+\frac{1}{2n}\Big)^{2n}\nearrow e$ as $n\rightarrow\infty$, $\Big(\frac{a_{n+1}}{a_n}\Big)^n\geq\frac1e$ for all $n\geq1$. Hence, by Frink's test, the series $\sum_na_n$ diverges.

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