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Have a look at this question:

Test whether the following series is conditionally convergent or absolutely convergent: $$ 1-\frac{1}{2}+\frac{1 \cdot 3}{2 \cdot 4}-\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6}+\ldots $$

My attempt: Since the given series is $$1+\sum_{n=1}^\infty (-1)^n\frac{1 \cdot 3\cdots 2n-1}{2\cdot 4 \cdots 2n}=\sum_{n=0}^\infty (-1)^na_n,~\text{where,}~a_0=1,~a_n=\frac{1 \cdot 3\cdots 2n-1}{2\cdot 4 \cdots 2n},~n\geq1.$$ See that $\displaystyle \frac{a_{n+1}}{a_n}=\frac{2n-1}{2n+1}<1 \Rightarrow a_{n+1}<a_{n}$ for all $n$.

Now, I have to verify $$\lim_{n \to \infty}a_n=\lim_{n \to \infty}\frac{1 \cdot 3\cdots 2n-1}{2\cdot 4 \cdots 2n}=0,$$ to apply alternating series test.

Observe that $$a_n=a_n \cdot \frac{2\cdot 4 \cdots 2n}{2\cdot 4 \cdots 2n}=\frac{1 \cdot 2\cdot 3\cdots 2n}{(2\cdot 4 \cdots 2n)^2}=\frac{(2n)!}{4^n(n!)^2}$$

Thereafter, I tried to make a sandwich $$\frac{(2n)!}{4^n(2n)!^2}\leq \frac{(2n)!}{4^n(n!)^2}\leq \frac{(2n)!}{4^nn!}$$ $$\Rightarrow \frac{1}{4^n(2n)!}\leq \frac{(2n)!}{4^n(n!)^2}\leq \frac{(2n)!}{4^nn!}$$

How to show right hand sequence goes to $0$ as $n \to \infty$, although I am feeling so.

Also, what about the absolute convergence as I failed to conclude by ratio test.

Messi Lio
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1 Answers1

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We have for $k\ge1$:

$$\begin{aligned} &(k+1)(2k-1)^2=4k^3-3k+1<4k^3,\\ &k(2k-1)^2=4k^3-4k^2+k>4k^3-4k^2=4k^2(k-1). \end{aligned} $$

This means: $$ \frac{k-1}k<\frac{(2k-1)^2}{(2k)^2}<\frac{k}{k+1}. $$

From $$ \left[\frac12\prod_{k=2}^n\frac{2k-1}{2k}\right]^2<\frac14\prod_{k=2}^n\frac{k}{k+1}=\frac1{2n+2}, $$ and $$ \left[\frac12\prod_{k=2}^n\frac{2k-1}{2k}\right]^2>\frac14\prod_{k=2}^n\frac{k-1}{k}=\frac1{4n}, $$ valid for $n>1$ one concludes: $$ \frac1{\sqrt{4n}}<\prod_{k=1}^n\frac{2k-1}{2k}<\frac1{\sqrt{2n+2}}. $$

This is much weaker than the Stirling estimate but suffices to solve your problem. The estimate can be sharpened if instead of $\frac12$ the factor $\prod_{k=1}^m\frac{2k-1}{2k}$ is splitted apart. Then the corresponding inequality is valid for $n>m$.

user
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