# Convergence and divergence of an infinite series

The series is $$1 + \frac{1}{2}.\frac{x^2}{4} + \frac{1\cdot3\cdot5}{2\cdot4\cdot6}.\frac{x^4}{8} + \frac{1\cdot3\cdot5\cdot7\cdot9}{2\cdot4\cdot6\cdot8\cdot10}.\frac{x^6}{12}+... , x\gt0$$ I just stuck over the nth term finding and once I get nth term than I can do different series test but here I am unable to find the nth term of the given series please help me out of this. The question is different as it contains x terms and its nth term will be totally different from marked as duplicate question

• Did you mean $1\cdot3\cdot5$ instead of $1.3.5$? – Kyky Dec 14 '17 at 5:34
• its multiplication of 1,3,5 = 1x3x5 =15 – KD. Dec 14 '17 at 5:35
• – erfink Dec 14 '17 at 8:30
• not duplicate as here every terms are different – KD. Dec 14 '17 at 9:23
• @KD.: the work shown in your answer provides context for your question. To keep the question from being closed, you should either move that work into the question or mention in the question that you have written an answer. – robjohn Dec 14 '17 at 14:16

Hint 1: Since $(4n-2)^2\gt(4n-1)(4n-3)$, we have \begin{align} \frac{1\cdot3\cdots(4n-3)}{2\cdot4\cdots(4n-2)} &\le\sqrt{\frac{1\cdot3\cdots(4n-3)}{3\cdot5\cdots(4n-1)}\cdot\frac{1\cdot3\cdots(4n-3)}{1\cdot3\cdots(4n-3)}}\\ &=\sqrt{\frac1{4n-1}} \end{align} Hint 2: Since $(4n-3)^2\gt(4n-2)(4n-4)$, we have \begin{align} \frac12\cdot\frac{3\cdot5\cdots(4n-3)}{4\cdot6\cdots(4n-2)} &\ge\frac12\sqrt{\frac{4\cdot6\cdots(4n-2)}{4\cdot6\cdots(4n-2)}\cdot\frac{2\cdot4\cdots(4n-4)}{4\cdot6\cdots(4n-2)}}\\ &=\frac12\sqrt{\frac1{2n-1}} \end{align}

Thus, $$\frac12\sqrt{\frac1{2n-1}}\,\frac{x^{2n}}{4n}\le\frac{1\cdot3\cdots(4n-3)}{2\cdot4\cdots(4n-2)}\frac{x^{2n}}{4n}\le\sqrt{\frac1{4n-1}}\,\frac{x^{2n}}{4n}$$ and therefore, since $4n-1\ge3n$ for $n\ge1$, we have $$\bbox[5px,border:2px solid #C0A000]{\frac{x^{2n}}{8\sqrt2\,n^{3/2}}\le\frac{1\cdot3\cdots(4n-3)}{2\cdot4\cdots(4n-2)}\frac{x^{2n}}{4n}\le\frac{x^{2n}}{4\sqrt3\,n^{3/2}}}$$

Hints:

The coefficient of the $n^\text{th}$ term can be written as: $$\frac{1\times 3 \times 5 … (1+4(n-2))}{2\times 4\times 6 …(2+4(n-2))}$$ $$=\frac{1\times 2 \times 3 … (1+4(n-2))\times (2+4(n-2))}{[2\times 4 \times … (2+4(n-2))]^2}$$ $$=\frac{(4n-6)!}{[(2n-3)!(2^{2n-3})]^2}$$ $$=\frac{(4n-6)!}{2^{4n-6}(2n-3)!^2}$$

The power of $x$ in the $n^{\text{th}}$ term with the number in the denominator can be written as: $$\frac{x^{2n-2}}{4n-4}$$

Thus, the $n^{\text{th}}$ term is: $$\frac{(4n-6)!}{2^{4n-6}(2n-3)!^2}\frac{x^{2n-2}}{4n-4}$$

• I guess $a_{n} = \frac{(2n-1)!!}{(2n+2)(2n)!!}$ – openspace Dec 14 '17 at 5:48
• I posted my answer what i actually getting after doing this but still not getting correct answer – KD. Dec 14 '17 at 7:12

$$\frac{1.2.2.x^2}{2.2.4} + \frac{1.3.5.2.4.6.x^4}{(2.4.6)^2.8}+\frac{1.3.5.7.9.2.4.6.10.8.x^6}{(2.4.6.8.10)^2.12}+...$$
so on sloving an = $$\text{an}=\frac{x^{2n}}{2^{3n}.n.(2n-1)!}$$ now applying d`alemebert ratio test $$\frac{\text{lim}}{n->∞} \frac{a[n]}{a[n+1]}=\frac{x^2.n}{8(n+1).(2n+1)}$$ which will be 0 so it converges but actually the correct answer is $$x^2<=1$$ $$\text{converge else diverge}$$

• The general term is $\binom{4n-2}{2n-1}\frac{x^{2n}}{2^{4n}n}$, but it takes work from getting that information to finding where the series converges. – robjohn Dec 14 '17 at 14:29

Consider $\displaystyle a_{n}x^{2n+2} = \frac{(2n-1)!!}{(2n+2)(2n)!!}x^{2n+2}$. By Cauchy-test : $\displaystyle \lim_{n\to \infty}\left(\frac{x^{2n+2}(2n-1)!!}{(2n+2)(2n)!!}\right)^{1/n} = \lim_{n\to \infty}\left(\frac{x^{2n+2}(2n-1)!}{4^{n}(n!)^{2}}\right)^{1/n}$, using Stirling's approximation ($n!$~$\frac{\sqrt{2\pi n}n^{n}}{e^{n}}$) , you will get your radius of convergency.

• I posted my answer what i actually getting after doing this but still not getting correct answer – KD. Dec 14 '17 at 7:12

Hint:

\begin{align} \frac{1\cdot 3 \cdot 5 \cdot 7 \cdot 9}{2\cdot 4 \cdot 6 \cdot8\cdot10}.\frac{x^6}{12} &=\frac{(1\cdot 3 \cdot 5 \cdot 7 \cdot 9)(2\cdot 4 \cdot 6 \cdot8\cdot10)}{(2\cdot 4 \cdot 6 \cdot8\cdot10)^2}.\frac{x^6}{12} \\ &=\frac{10!}{2^{10}(1\cdot 2 \cdot 3 \cdot4\cdot 5)^2} \frac{x^6}{12}\\ &=\frac{1}{2^{10}} \cdot \binom{10}{5} \frac{x^6}{2(6)}\\ &= \frac{1}{2^{2\cdot (6-1)}} \cdot \binom{2\cdot(6-1)}{6-1} \frac{x^6}{2(6)} \end{align}

Edit:

$$\frac{1}{2^{2(n-1)}}\frac{(2n-2)!}{((n-1)!)^2}\frac{x^{2n}}{2n}$$

• I posted my answer what i actually getting after doing this but still not getting correct answer – KD. Dec 14 '17 at 7:12
• I don't think your $a_n$ term is correct. – Siong Thye Goh Dec 14 '17 at 7:18

Notice \begin{align} \frac{1}{2\cdot 4}x^2 &= \frac{2!}{(2^1 1!)^2(2 \cdot 2)} x^2 = \frac12 \binom{2}{1} \int_0^x \left(\frac{t}{4}\right)^1 dt \\ \frac{1\cdot 3\cdot 5}{2\cdot 4 \cdot 6 \cdot 8}x^4 &= \frac{6!}{(2^3 3!)^2 (2\cdot 4)} x^4 = \frac12 \binom{6}{3} \int_0^x \left(\frac{t}{4}\right)^3 dt\\ \frac{1\cdot 3\cdot 5\cdot 7\cdot 9}{2\cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12}x^6 &= \frac{10!}{(2^5 5!)^2 (2\cdot 6)} x^6 = \frac12 \binom{10}{5} \int_0^x \left(\frac{t}{4}\right)^5 dt \end{align} So aside from the first constant term $1$, the $k^{th}$ non-constant term has the form $$a_k = \frac12\binom{2\ell}{\ell}\frac{x^{\ell+1}}{2^{2\ell+1}(\ell+1)} = \frac12\binom{2\ell}{\ell}\int_0^x \left(\frac{t}{4}\right)^\ell dt$$ where $\ell = 2k-1$, the $k^{th}$ odd number.

If one compute the ratio of successive terms of $a_k$, one find

$$\frac{a_{k+1}}{a_k} = \frac{\binom{4k+2}{2k+1}(2k)}{\binom{4k-2}{2k-1}(2k+2)} \frac{x^4}{2^4} = \frac{(4k+2)(4k+1)(4k)(4k-1)}{(2k+1)^2(2k)(2k+2)}\frac{x^4}{2^4} = \frac{\left(1-\frac{1}{4k}\right)^2}{\left(1 + \frac{1}{2k}\right)\left(1+\frac1k\right)} x^4$$ Since this ratio converges to $x^4$ as $k \to \infty$, the radius of convergence of the series is $1$.

In fact, for $|x| < 1$, the series has following representation $$1 + \frac12\sum_{\ell\text{ odd}}\binom{2\ell}{\ell}\int_0^x \left(\frac{t}{4}\right)^\ell dt = 1 + \frac14\sum_{\ell=0}^\infty \binom{2\ell}{\ell} (1 - (-1)^{\ell})\int_0^x \left(\frac{t}{4}\right)^\ell dt$$ For $|z| < \frac14$, we have following identity:

$$\sum_{\ell=0}^\infty \binom{2\ell}{\ell} z^\ell = \frac{1}{\sqrt{1-4z}}$$

From this, we find for $|x| < 1$, the series evaluates to $$1 + \frac14 \int_0^x \left(\frac{1}{\sqrt{1-t}} - \frac{1}{\sqrt{1+t}}\right) dt = 2 - \frac12\left( \sqrt{1+x} + \sqrt{1-x}\right)$$