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I have the following series: $$\sum_{k=1}^\infty \frac{(2k)!}{4^k(k!)^2}$$ Using ratio test I find: $\lim_{k\to \infty}\frac{a_{n+1}}{a_n}=\lim_{k\to \infty}\frac{2k+1}{2(k+1)}=1$ I'm trying to form $1-\frac{a}{k}$ since ratio test doesn't help.

$\frac{2k+1}{2(k+1)}=\frac{2k+2-1}{2k+2}=1-\frac{1}{2k+2}=1-\frac{1}{2k}\frac{1}{1+1/k}=1-\frac{1}{2k}(1+1/k)^{-1}$ With binomial expansion we have : $1-\frac{1}{2k}(1-1/k+1/k^2 -1/k^3+...)$ How do I proceed now? I think our teacher ignored the whole parenthesis somehow and said this : $1-\frac{1}{2k}(1-1/k+1/k^2 -1/k^3+...)\sim1-\frac{1}{2k}$but it doesn't make sense to me.

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    $$1 - \frac{1}{2(k+1)} = 1 - \frac{1}{2k} + \frac{1}{2k(k+1)}$$ If you have a formulation of Raabe's test using $1 - \frac{a}{k} + O(k^{-(1+\varepsilon)})$ (or $o(k^{-1})$), that suffices. What formulation of the test have you? – 2017-01-22
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    I only have 1-a/k and if a<1 the series diverges. If a>1 it converges. – 2017-01-22
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    But one practically never has $\frac{a_{k+1}}{a_k} = 1 - \frac{a}{k}$ with an $a$ independent of $k$. So that wouldn't be a useful form of the test. Can you quote how _exactly_ the test is given in your course? – 2017-01-22
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    We were only given this form to use when the ratio test fails and this single exercise. And it could also be in the test I'm having tomorrow. I could ignore the parenthesis just like he did but I wanna know how he's allowed to do what he did. I'm sure there are better solutions but we were only taught this one. – 2017-01-22
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    Please, what was the exact formulation of the test you were given? – 2017-01-22
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    While there's chat about other ways to show divergence going on, how about this one: For $\lvert x\rvert < 1$ we have $$(1 - x)^{-\frac{1}{2}} = \sum_{k = 0}^{\infty} (-1)^k\binom{-\frac{1}{2}}{k}x^k = \sum_{k = 0}^{\infty} \frac{(2k)!}{4^k(k!)^2}x^k.$$ Now let $x \to 1$. – 2017-01-22
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    I'm sorry, maybe I don't understand because of my English. By formulation of the test I thought you meant Raabe's test as it was given by our teacher .What do you mean ? – 2017-01-22
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    Yes, I would like to know the exact way your teacher gave Raabe's test. No shortcuts, the exact words and symbols. – 2017-01-22
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    Maybe I got it wrong and it's not Raabe's test. He told us it's the generalised version of the ratio test. If the limit of the ratio test is 1 then if a(n+1)/a(n) ∼ 1-λ/n then: if λ>1 the series converges , if λ<1 the series diverges – 2017-01-22
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    Aha, we have a $\sim$, not an $=$. Now the question is how that $\sim$ is to be interpreted. Raabe's test interprets it as $$\lambda = \lim_{n\to\infty} n\biggl(1 - \frac{a_{n+1}}{a_n}\biggr).$$ – 2017-01-22
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    Well maybe he's trying to make it less complicated to us. I found another exercise where he did the same thing . He ignored all the k with a power of 2 and above and used ∼ again. I'll have to publish another question about that. Thank you sir! – 2017-01-22

2 Answers 2

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Using Raabe's test, we have

$$\begin{align} k\left(\frac{a_k}{a_{k+1}}-1\right)&=k\left(\frac{4(2k)!((k+1)!)^2}{(k!)^2(2k+2)!}-1\right)\\\\ &=k\left(\frac{2(k+1)}{2k+1}-1\right)\\\\ &=\frac{k}{2k+1}\to \frac12\,\,\,\text{as}\,\,k\to \infty \end{align}$$

Inasmuch as the limit is less than $1$, the tests guarantees that the series diverges.

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It is a diverging series since $$ \frac{1}{4^k}\binom{2k}{k}\approx\frac{1}{\sqrt{\pi k}}.\tag{1} $$

You may notice that the square of the LHS is given by

$$\prod_{j=1}^{k}\left(1-\frac{1}{2j}\right)^2 = \frac{1}{4}\prod_{j=2}^{k}\left(1-\frac{1}{j}\right)\prod_{j=2}^{k}\left(1-\frac{1}{(2j-1)^2}\right)^{-1} \\=\frac{1}{4k}\prod_{j=2}^{k}\left(1-\frac{1}{(2j-1)^2}\right)^{-1}\tag{2}$$ and the infinite product $\prod_{j\geq 2}\left(1-\frac{1}{(2j-1)^2}\right)^{-1}$ is convergent to $\frac{4}{\pi}$.

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    We are not doing infinte products in my class so I have trouble understanding what you did there.I could look it up but unfortunately we have to use the binomial expansion way to solve it. Thank you though. – 2017-01-22
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    @JohnKatsantas: it is not essential to know the exact value of the mentioned infinite product (Wallis product). The key part is that by squaring $\frac{1}{4^k}\binom{2k}{k}$ you get something that behaves like $\frac{C}{\sqrt{k}}$, and that is enough to prove divergence. – 2017-01-22