- #1

mathmari

Gold Member

MHB

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I want to check the convergence for the below series.

- $\displaystyle{\sum_{n=1}^{+\infty}\frac{\left (n!\right )^2}{\left (2n+1\right )!}4^n}$

Let $\displaystyle{a_n=\frac{\left (n!\right )^2}{\left (2n+1\right )!}\cdot 4^n}$.

Then we have that \begin{align*}a_{n+1}&=\frac{\left ((n+1)!\right )^2}{\left (2(n+1)+1\right )!}\cdot 4^{n+1}=\frac{(n!)^2(n+1)^2}{\left (2n+3\right )!}\cdot 4^{n}\cdot 4=\frac{(n!)^2(n+1)^2}{\left (2n+1\right )!(2n+2)(2n+3)}\cdot 4^{n}\cdot 4 \\ & =\frac{(n!)^2(n+1)^2}{\left (2n+1\right )!2(n+1)(2n+3)}\cdot 4^{n}\cdot 4 =\frac{(n!)^2(n+1)}{\left (2n+1\right )!(2n+3)}\cdot 4^{n}\cdot 2\end{align*}

So we get \begin{equation*}\frac{a_{n+1}}{a_n}=\frac{\frac{(n!)^2(n+1)}{\left (2n+1\right )!(2n+3)}\cdot 4^{n}\cdot 2}{\frac{\left (n!\right )^2}{\left (2n+1\right )!}\cdot 4^n}=\frac{(n!)^2(n+1)\left (2n+1\right )!}{\left (2n+1\right )!(2n+3)(n!)^2}\cdot 2=\frac{(n+1)}{(2n+3)}\cdot 2=\frac{2n+2}{2n+3}\end{equation*}

It holds that $\displaystyle{\left |\frac{a_{n+1}}{a_n}\right |=\frac{2n+2}{2n+3}<1}$.

For the ratio test we have that the limit has to be less than $1$, right? So do we have to apply an other test?

- $\displaystyle{\sum_{n=1}^{+\infty}\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{2\cdot 4\cdot 6\cdot \ldots \cdot 2n}}$

Let $a_n=\prod_{i=1}^n\frac{2i-1}{2i}$ then $a_{n+1}=\prod_{i=1}^{n+1}\frac{2i-1}{2(i+1)}$.

So we get that $\frac{a_{n+1}}{a_n}=\frac{2(n+1)-1}{2(n+1+1)}=\frac{2n+1}{2n+4}<1$

Again the ration test doesn;t help us, does it?

- $\displaystyle{\sum_{n=1}^{+\infty}\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{2\cdot 4\cdot 6\cdot \ldots \cdot 2n\cdot (2n+2)}}$

Let $a_n=\prod_{i=1}^n\frac{2i-1}{2i+2}$ then $a_{n+1}=\prod_{i=1}^{n+1}\frac{2i-1}{2i+2}$.

So we get that $\frac{a_{n+1}}{a_n}=\frac{2(n+1)-1}{2(n+1)+2}<1$.

The same here. So do we have to fins inequalities at each case or to use an other test? :unsure: