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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

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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 $$ \bbox[5px,border:2px solid #C0A000]{\frac{x^{2n}}{8\sqrt2\,n^{3/2}}\le\frac{1\cdot3\cdots(4n-3)}{2\cdot4\cdots(4n-2)}\frac{x^{2n}}{4n}\le\frac{x^{2n}}{4\sqrt3\,n^{3/2}}} $$

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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: $$\frac{(4n-6)!}{2^{4n-6}(2n-3)!^2}\frac{x^{2n-2}}{4n-4}$$

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  • $\begingroup$ I guess $a_{n} = \frac{(2n-1)!!}{(2n+2)(2n)!!}$ $\endgroup$ – openspace Dec 14 '17 at 5:48
  • $\begingroup$ I posted my answer what i actually getting after doing this but still not getting correct answer $\endgroup$ – KD. Dec 14 '17 at 7:12
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$$\frac{1.2.2.x^2}{2.2.4} + \frac{1.3.5.2.4.6.x^4}{(2.4.6)^2.8}+\frac{1.3.5.7.9.2.4.6.10.8.x^6}{(2.4.6.8.10)^2.12}+...$$
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 $$x^2<=1$$ $$ \text{converge else diverge}$$

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  • $\begingroup$ 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. $\endgroup$ – robjohn Dec 14 '17 at 14:29
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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.

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

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

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  • $\begingroup$ I posted my answer what i actually getting after doing this but still not getting correct answer $\endgroup$ – KD. Dec 14 '17 at 7:12
  • $\begingroup$ I don't think your $a_n$ term is correct. $\endgroup$ – Siong Thye Goh Dec 14 '17 at 7:18
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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 $$1 + \frac14 \int_0^x \left(\frac{1}{\sqrt{1-t}} - \frac{1}{\sqrt{1+t}}\right) dt = 2 - \frac12\left( \sqrt{1+x} + \sqrt{1-x}\right)$$

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