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2018 Paper 2 Q5
D: 1600.0 B: 1505.3
Taylor series binomial expansion integration logarithmic series exponential series term-by-term integration substitution infinite series definite integral Frullani integral

In this question, you should ignore issues of convergence.

  1. Write down the binomial expansion, for \(\vert x \vert<1\,\), of \(\;\dfrac{1}{1+x}\,\) and deduce that %. By considering %$ %\displaystyle \int \frac 1 {1+x} \, \d x %\,, %$ %show that \[ \displaystyle \ln (1+x) = -\sum_{n=1}^\infty \frac {(-x)^n}n \, \] for \(\vert x \vert <1 \,\).
  2. Write down the series expansion in powers of \(x\) of \(\displaystyle \e^{-ax}\,\). Use this expansion to show that \[ \int_0^\infty \frac {\left(1- \e^{-ax}\right)\e^{-x}}x \,\d x = \ln(1+a) \ \ \ \ \ \ \ (\vert a \vert <1)\,. \]
  3. Deduce the value of \[ \int_0^1 \frac{x^p - x^q}{\ln x} \, \d x \ \ \ \ \ \ (\vert p\vert <1, \ \vert q\vert <1) \,. \]

Solution:

  1. \begin{align*} && \frac1{1+x} &= 1 - x+ x^2 - x^3+ \cdots \\ \Rightarrow && \int_0^x \frac{1}{1+t} \d t &= \int_0^x \sum_{n=0}^{\infty} (-t)^n \d t \\ &&&= \left [\sum_{n=0}^{\infty} -\frac{(-t)^{n+1}}{n+1} \right]_0^x \\ \Rightarrow &&\ln(1+x)&=- \sum_{n=1}^\infty \frac{(-x)^n}{n} \end{align*}
  2. \begin{align*} && e^{-ax} &= \sum_{n=0}^\infty \frac{(-a)^n}{n!} x^n \\ \Rightarrow && \int_0^{\infty} \frac{1}{x} \left (1-e^{-ax} \right)e^{-x} \d x &= \int_0^{\infty} \frac{1}{x} \left (-\sum_{n=1}^\infty \frac{(-a)^n}{n!}x^n \right)e^{-x} \d x \\ &&&= -\int_0^{\infty} \sum_{n=1}^\infty \frac{(-a)^n}{n!} x^{n-1} e^{-x} \d x \\ &&&= -\sum_{n=1}^\infty \frac{(-a)^n}{n!} \int_0^{\infty} x^{n-1} e^{-x} \d x \\ &&&= -\sum_{n=1}^\infty \frac{(-a)^n}{n!} (n-1)! \\ &&&= -\sum_{n=1}^\infty \frac{(-a)^n}{n} \\ &&&= \ln (1+a) \end{align*}
  3. \begin{align*} && \int_0^1 \frac{x^p - x^q}{\ln x} \, \d x &= \int_0^1 \frac{x^p(1 - x^{q-p})}{\ln x} \, \d x \\ e^{-u} = x, \d x = -e^{-u} \d u: &&&=\int_{u=\infty}^{0} \frac{e^{-pu}-e^{-qu}}{-u} (-e^{-u})\d u \\ &&&= \int_0^\infty \frac{e^{-u}(e^{-qu}-e^{-pu})}{u} \d u \\ &&&= \int_0^\infty \frac{e^{-(1+q)u}(1-e^{-(p-q)u})}{u} \d u \\ v = (1+q)u, \d v = (1+q) \d u: &&&=\int_0^{\infty} \frac{e^{-v}(1-e^{-\left(\frac{p-q}{1+q}\right)v}}{v}\d v \\ &&&= \ln \left(1 + \frac{p-q}{1+q} \right) \\ &&&= \ln \left ( \frac{1+p}{1+q} \right) \end{align*}

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2014 Paper 3 Q1
D: 1700.0 B: 1542.2
roots of polynomials symmetric functions Newton's sums logarithmic series power sums series expansion proof by counterexample

Let \(a\), \(b\) and \(c\) be real numbers such that \(a+b+c=0\) and let \[(1+ax)(1+bx)(1+cx) = 1+qx^2 +rx^3\,\] for all real \(x\). Show that \(q = bc+ca+ab\) and \(r= abc\).

  1. Show that the coefficient of \(x^n\) in the series expansion (in ascending powers of \(x\)) of \(\ln (1+qx^2+rx^3)\) is \((-1)^{n+1} S_n\) where \[S_n = \frac{a^n+b^n+c^n}{n} \,, \ \ \ \ \ \ \ \ (n\ge1).\]
  2. Find, in terms of \(q\) and \(r\), the coefficients of \(x^2\), \(x^3\) and \(x^5\) in the series expansion (in ascending powers of \(x\)) of \(\ln (1+qx^2+rx^3)\) and hence show that \(S_2S_3 =S_5\).
  3. Show that \(S_2S_5 =S_7\).
  4. Give a proof of, or find a counterexample to, the claim that \(S_2S_7=S_9\).

Solution: \begin{align*} (1+ax)(1+bx)(1+cx) &= (1+(a+b)x+abx^2)(1+cx) \\ &= 1+(a+b+c)x+(ab+bc+ca)x^2+abcx^3 \end{align*} Therefore by comparing coefficients, \(q = bc + ca + ab\) and \(r = abc\) as required.

  1. \begin{align*} \ln (1+qx^2 + rx^3) &= \ln(1+ax) + \ln(1+bx)+\ln(1+cx) \\ &= -\sum_{n=1}^{\infty} \frac{(-ax)^n}{n}-\sum_{n=1}^{\infty} \frac{(-bx)^n}{n}-\sum_{n=1}^{\infty} \frac{(-cx)^n}{n} \\ &= \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(a^n+b^n+c^n)}{n} x^n \\ &= \sum_{n=1}^{\infty} (-1)^{n+1} S_n x^n \\ \end{align*}
  2. \begin{align*} \ln (1 + qx^2 + rx^3) &= (qx^2+rx^3) -\frac{(qx^2+rx^3)^2}{2} + O(x^6) \\ &= qx^2 + rx^3 - \frac12 q^2 x^4 - qr x^5 + O(x^6) \\ \end{align*} Comparing coefficients we see that \(S_2 = -q\) and \(S_3 = r\), we also must have \(S_5 = -qr = S_2S_3\) as required.
  3. \begin{align*} \ln (1 + qx^2 + rx^3) &= (qx^2+rx^3) -\frac{(qx^2+rx^3)^2}{2} +\frac{(qx^2+rx^3)^3}{3}+ O(x^8) \\ &= qx^2 + rx^3 - \frac12 q^2 x^4 - qr x^5 + \frac12 rx^6 + \frac13 q^3 x^6 + q^2r x^7 + O(x^8) \\ &= qx^2 + rx^3 - \frac12 q^2 x^4 - qr x^5 + \left ( \frac12 r+ \frac13 q^3 \right)x^6 + q^2r x^7 \end{align*} Comparing coefficients we see that \(S_2 = -q\) and \(S_5 =-qr\), we also must have \(S_7 = q^2r = S_2S_5\) as required.
  4. Let \(a = b = 1, c = -2\), then \(S_2 = \frac{1^2+1^2 + (-2)^2}{2} = 3, S_7 = \frac{1^2+1^2+(-2)^7}{7} = -18, S_9 = \frac{1^1+1^2+(-2)^9}{9} = \frac{2 - 512}{9} \neq 3 \cdot (-18)\)

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2012 Paper 2 Q4
D: 1600.0 B: 1500.0
Taylor series logarithmic series inequality ln(1+x) Euler's number e convergence series comparison limits

In this question, you may assume that the infinite series \[ \ln(1+x) = x-\frac{x^2}2 + \frac{x^3}{3} -\frac {x^4}4 +\cdots + (-1)^{n+1} \frac {x^n}{n} + \cdots \] is valid for \(\vert x \vert <1\).

  1. Let \(n\) be an integer greater than 1. Show that, for any positive integer \(k\), \[ \frac1{(k+1)n^{k+1}} < \frac1{kn^{k}}\,. \] Hence show that \(\displaystyle \ln\! \left(1+\frac1n\right) <\frac1n\,\). Deduce that \[ \left(1+\frac1n\right)^{\!n}<\e\,. \]
  2. Show, using an expansion in powers of \(\dfrac1y\,\), that $ \displaystyle \ln \! \left(\frac{2y+1}{2y-1}\right) > \frac 1y %= \sum _{r=0}^\infty \frac 1{(2r+1)(2y)^{2r}}\,. \( for \)y>\frac12$. Deduce that, for any positive integer \(n\), \[ \e < \left(1+\frac1n\right)^{\! n+\frac12}\,. \]
  3. Use parts (i) and (ii) to show that as \(n\to\infty\) \[ \left(1+\frac1n\right)^{\!n} \to \e\,. \]

Solution:

  1. Since \(k \geq 1\) we have \(n^{k+1} > n^k\) and \((k+1) > k\), therefore \((k+1)n^{k+1} >kn^k \Rightarrow \frac{1}{(k+1)n^{k+1}} < \frac{1}{kn^k}\) \begin{align*} && \ln \left ( 1 + \frac1n \right) &= \frac1n -\frac{1}{2n^2} + \frac{1}{3n^3} - \frac{1}{4n^4} + \cdots \\ &&&= \frac1n - \underbrace{\left (\frac{1}{2n^2}-\frac{1}{3n^3} \right)}_{>0}- \underbrace{\left (\frac{1}{4n^4}-\frac{1}{5n^5} \right)}_{>0} - \cdot \\ &&&< \frac1n \\ \\ \Rightarrow && n \ln \left ( 1 + \frac1n \right) &< 1 \\ \Rightarrow && \ln \left ( \left ( 1 + \frac1n \right)^n \right) &< 1 \\ \Rightarrow && \left ( 1 + \frac1n \right)^n &< e \end{align*}
  2. \(\,\) \begin{align*} &&\ln \left(\frac{2y+1}{2y-1}\right) &= \ln \left (1 + \frac{1}{2y} \right)-\ln \left (1 - \frac{1}{2y} \right) \\ &&&= \frac{1}{2y} - \frac{1}{2(2y)^2} + \frac{1}{3(2y)^3} - \cdots - \left (-\frac{1}{2y} - \frac{1}{2(2y)^2} - \frac{1}{3(2y)^3} - \cdots \right) \\ &&&= \frac{1}{y} + \frac{2}{3(2y)^3} + \frac{2}{5(2y)^5} \\ &&&= \sum_{r=1}^{\infty} \frac{2}{(2r-1)(2y)^{2r-1}} \\ &&&> \frac1y \\ \\ \Rightarrow && \ln \left (1 + \frac{1}{y-\frac12} \right) &> \frac{1}{y} \\\Rightarrow && \ln \left (1 + \frac{1}{n} \right) &> \frac{1}{n+\frac12} \\ \Rightarrow &&(n+\tfrac12) \ln \left (1 + \frac{1}{n} \right) &> 1\\ \Rightarrow && \ln \left ( \left (1 + \frac{1}{n} \right)^{n+\tfrac12} \right) &> 1\\ \Rightarrow && \left (1 + \frac{1}{n} \right)^{n+\tfrac12} & > e \end{align*}
\item Since \(\left (1 + \frac1n \right)^n\) is both bounded above, and increasing, it must tend to some limit \(L\). \begin{align*} && \lim_{n \to \infty} \left (1 + \frac1n \right)^n && \leq e &\leq & \lim_{n \to \infty} \left (1 + \frac1n \right)^{n+\frac12} \\ \Rightarrow && \lim_{n \to \infty} \left (1 + \frac1n \right)^n && \leq e &\leq & \lim_{n \to \infty} \left (1 + \frac1n \right)^{n} \lim_{n \to \infty} \sqrt{1 + \frac1n} \\ \Rightarrow && \lim_{n \to \infty} \left (1 + \frac1n \right)^n && \leq e &\leq & \lim_{n \to \infty} \left (1 + \frac1n \right)^{n} \\ \end{align*} And therefore equality must hold.

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2012 Paper 3 Q2
D: 1700.0 B: 1516.0
sequences infinite product telescoping logarithmic series partial fractions geometric series convergence proof by induction

In this question, \(\vert x \vert <1\) and you may ignore issues of convergence.

  1. Simplify \[ (1-x)(1+x)(1+x^2)(1+x^4) \cdots (1+x^{2^n})\,, \] where \(n\) is a positive integer, and deduce that \[ \frac1{1-x} = (1+x)(1+x^2)(1+x^4) \cdots (1+x^{2^n}) + \frac {x^{2^{n+1}}}{1-x}\,. \] Deduce further that \[ \ln(1-x) = - \sum_{r=0}^\infty \ln \left (1+ x ^{2^r}\right) \,, \] and hence that \[ \frac1 {1-x} = \frac 1 {1+x} + \frac {2x}{1+x^2} + \frac {4x^3}{1+x^4} +\cdots\,. \]
  2. Show that \[ \frac{1+2x}{1+x+x^2} = \frac{1-2x}{1-x+x^2} + \frac{2x-4x^3}{1-x^2+x^4} + \frac {4x^3-8x^7}{1-x^4+x^8} + \cdots\,. \]

Solution:

  1. \begin{align*} (1-x)&(1+x)(1+x^2)(1+x^4) \cdots (1+x^{2^n}) \\ &= (1-x^2)(1+x^2)(1+x^4) \cdots (1+x^{2^n}) \\ &= (1-x^4)(1+x^4) \cdots (1+x^{2^n}) \\ &= 1-x^{2^{n+1}} \\ \end{align*} Therefore, \begin{align*} && \frac{1}{1-x} - \frac{x^{2^{n+1}}}{1-x} &= (1+x)(1+x^2)\cdots(1+x^{2^n}) \\ \Rightarrow && \frac{1}{1-x} &=(1+x)(1+x^2)\cdots(1+x^{2^n})+ \frac{x^{2^{n+1}}}{1-x} \\ \Rightarrow && -\ln (1-x) &= \sum_{r=0}^{\infty} \ln (1+x^{2^r}) \\ \Rightarrow && \ln(1-x) &= - \sum_{r=0}^{\infty} \ln (1+x^{2^r}) \\ \underbrace{\Rightarrow}_{\frac{\d}{\d x}} && \frac{1}{1-x} &= \sum_{r=0}^{\infty} \frac{2^r x^{2^r-1}}{1+x^{2^r}} \\ &&&= \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \cdots \end{align*}
  2. Consider \begin{align*}(1+x+x^2)&(1-x+x^2)(1-x^2+x^4)\cdots(1-x^{2^n}+x^{2^{n+1}}) \\ &= (1+x^2 + x^4)(1-x^2+x^4) \cdots (1-x^{2^n}+x^{2^{n+1}}) \\ &= (1-x^{2^{n+1}}+x^{2^{n+2}}) \\ \end{align*} Therefore, \begin{align*} && \frac{1}{1+x+x^2} &= (1-x+x^2)(1-x^2+x^4)\cdots(1-x^{2^n}+x^{2^{n+1}}) + \frac{x^{2^{n+1}}}{1+x+x^2} -\frac{x^{2^{n+2}}}{1+x+x^2} \\ \Rightarrow && -\ln(1+x+x^2) &= \sum_{r=0}^\infty \ln (1 - x^{2^r}+x^{2^{r+1}}) \\ \underbrace{\Rightarrow}_{\frac{\d}{\d x}} && -\frac{1+2x}{1+x+x^2} &= \sum_{r=0}^{\infty} \frac{ -2^r x^{2^r-1}+2^{r+1}x^{2^{r+1}-1}}{1 - x^{2^r}+x^{2^{r+1}}} \\ &&&= \frac{-1+2x}{1-x+x^2}+\frac{-2x+4x^3}{1-x^2+x^4} + \frac{-4x^3+8x^7}{1-x^4+x^8} + \cdots \end{align*} Which is the desired result when we multiply both sides by \(-1\)

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1993 Paper 3 Q4
D: 1700.0 B: 1500.0
sequences infinite series summation power series partial fractions geometric series logarithmic series telescoping series differentiation of series

Sum the following infinite series.

  1. \[ 1 + \frac13 \bigg({\frac12}\bigg)^2 +\frac15\bigg(\frac12\bigg)^4 + \cdots + \frac{1}{2n+1} \bigg(\frac12\bigg)^{2n} + \cdots \] .
  2. \[ 2 -x -x^3 +2x^4 - \cdots + 2x^{4k} - x^{4k+1} - x^{4k+3} +\cdots \] where \(|x| < 1\).
  3. \[ \sum _{r=2}^\infty {r\, 2^{r-2} \over 3^{r-1} } \].
  4. \[ \sum_{r=2}^\infty {2 \over r(r^2-1) } \].

Solution:

  1. \begin{align*} && \sum_{i=0}^{\infty} x^{2i+1}&= \frac{x}{1-x^2} \\ \Rightarrow &&&=\frac12 \left ( \frac{1}{1-x} - \frac{1}{1+x} \right) \\ \underbrace{\Rightarrow}_{\int} && \sum_{i=0}^{\infty} \frac{1}{2i+1} x^{2i+2} &= \frac12 \left ( -\ln (1-x) - \ln(1+x) \right) \\ \underbrace{\Rightarrow}_{x = 1/2} && \sum_{i=0}^\infty \frac{1}{2i+1} \left (\frac12 \right)^{2i+2} &= -\frac12 \ln \frac12 - \frac12 \ln \frac32 \\ &&&= -\frac12 \ln \frac34 \\ &&\frac14\sum_{i=0}^\infty \frac{1}{2i+1} \left (\frac12 \right)^{2i}&= \frac12 \ln \frac43 \\ \Rightarrow&& S &= 2 \ln \frac43 \end{align*}
  2. \begin{align*} \sum_{k=0}^{\infty} \left (2x^{4k} - x^{4k+1} - x^{4k+3} \right) &= \sum_{k=0}^{\infty} \left (2 - x^{1} - x^{3} \right) x^{4k} \\ &= \frac{2-x-x^3}{1-x^4} \\ &= \frac{(1-x)(2+x+x^2)}{(1-x)(1+x+x^2+x^3)} \\ &= \frac{2+x+x^2}{1+x+x^2+x^3} \end{align*}
  3. \begin{align*} && \frac{1}{(1-x)^2} &= \sum_{r=0}^{\infty} r x^{r-1} \\ \Rightarrow && 9 &= \sum_{r=1}^{\infty} r \left ( \frac23 \right)^{r-1} \\ \Rightarrow && \sum_{r=2}^{\infty} r \left ( \frac{2^{r-2}}{3^{r-1}} \right) &= \frac12 \left ( 9 - 1 \right) \\ &&&= 4 \end{align*}
  4. \begin{align*} && \frac{2}{r(r^2-1)} &= \frac{1}{r-1} - \frac{2}{r} + \frac{1}{r+1} \\ \Rightarrow && \sum_{r=2}^{\infty} \left ( \frac{1}{r-1} - \frac{2}{r} + \frac{1}{r+1} \right) &= \sum_{r=2}^{\infty} \left ( \frac{1}{r-1} - \frac{1}{r} - \frac{1}{r} + \frac{1}{r+1} \right) \\ &&&= \sum_{r=2}^{\infty} \left ( \frac{1}{r-1} - \frac{1}{r} \right)-\sum_{r=2}^{\infty} \left ( \frac{1}{r} - \frac{1}{r+1} \right) \\ &&&= 1 - \frac12 \\ &&&= \frac12 \end{align*}

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1991 Paper 3 Q1
D: 1700.0 B: 1501.5
Taylor series partial fractions telescoping sum logarithmic series exponential series power series expansion geometric series

  1. Evaluate \[ \sum_{r=1}^{n}\frac{6}{r(r+1)(r+3)}. \]
  2. Expand \(\ln(1+x+x^{2}+x^{3})\) as a series in powers of \(x\), where \(\left|x\right|<1\), giving the first five non-zero terms and the general term.
  3. Expand \(\mathrm{e}^{x\ln(1+x)}\) as a series in powers of \(x\), where \(-1 < x\leqslant1\), as far as the term in \(x^{4}\).

Solution:

  1. \begin{align*} && \frac{6}{r(r+1)(r+3)} &= \frac{2}{r} - \frac{3}{r+1} + \frac{1}{r+3} \\ \Rightarrow && \sum_{r=1}^n \frac{6}{r(r+1)(r+3)} &= \sum_{r=1}^n \l \frac{2}{r} - \frac{3}{r+1} + \frac{1}{r+3} \r \\ &&&= \sum_{r=1}^n \frac{2}{r} - \sum_{r=1}^n \frac{3}{r+1} + \sum_{r=1}^n \frac{1}{r+3} \\ &&&= \sum_{r=1}^n \frac{2}{r} - \sum_{r=2}^{n+1} \frac{3}{r} + \sum_{r=3}^{n+2} \frac{1}{r} \\ &&& = \frac{2}{1} + \frac{2}{2} - \frac{3}{2} - \frac{3}{n+1} + \frac{1}{n+1} + \frac{1}{n+2} \\ &&& = \frac{3}{2} - \frac{2}{n+1} + \frac{1}{n+2} \end{align*}
  2. \begin{align*} && \ln (1 + x+ x^2 + x^3) &= \ln \l \frac{1-x^4}{1-x} \r \\ &&&= \ln (1-x^4) - \ln(1-x) \\ &&&= \sum_{k=1}^{\infty} -\frac{x^{4k}}{k} - \sum_{k=1}^{\infty} - \frac{x^k}{k} \\ &&&= x + \frac12x^2+\frac13x^3-\frac34x^4+\frac15x^5 + \cdots \\ &&&= \sum_{k=1}^{\infty}a_k x^k \end{align*} Where \(a_k = \frac{1}{k}\) if \(k \neq 0 \pmod{4}\) otherwise \(a_k = -\frac{3}{k}\) if \(k \equiv 0 \pmod{4}\)
  3. \begin{align*} \exp(x \ln (1+x) ) &= \exp\l x \l x-\frac12x^2+\frac13x^3-\cdots \r \r \\ &= \exp\l x^2-\frac12x^3+\frac13x^4 \r \\ &= 1 + \l x^2-\frac12x^3+\frac13x^4 \r + \frac12 \l x^2-\frac12x^3+\frac13x^4 \r^2 + \cdots \\ &= 1 + x^2-\frac12x^3+\frac13x^4 + \frac12x^4 + \cdots \\ &= 1 + x^2 -\frac12x^3+\frac56x^4+\cdots \end{align*}

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1989 Paper 3 Q9
D: 1700.0 B: 1516.0
sequences sum to infinity power series arithmetico-geometric series logarithmic series binomial series series manipulation

Obtain the sum to infinity of each of the following series.

  1. \(1{\displaystyle +\frac{2}{2}+\frac{3}{2^{2}}+\frac{4}{2^{3}}+\cdots+\frac{r}{2^{r-1}}+\cdots;}\)
  2. \(1{\displaystyle +\frac{1}{2}\times\frac{1}{2}+\frac{1}{3}\times\frac{1}{2^{2}}+\cdots+\frac{1}{r}\times\frac{1}{2^{r-1}}+\cdots;}\)
  3. \({\displaystyle \dfrac{1\times3}{2!}\times\frac{1}{3}+\frac{1\times3\times5}{3!}\frac{1}{3^{2}}+\cdots+\frac{1\times3\times\cdots\times(2k-1)}{k!}\times\frac{1}{3^{k-1}}+\cdots.}\)
[Questions of convergence need not be considered.]

Solution:

  1. \begin{align*} && \frac1{1-x} &= \sum_{r=0}^{\infty} x^r \\ \underbrace{\Rightarrow}_{\frac{\d}{\d x}} && \frac{1}{(1-x)^2} &= \sum_{r=0}^\infty rx^{r-1} \\ \underbrace{\Rightarrow}_{x = \frac12} && 4 &= \sum_{r=0}^{\infty} \frac{r}{2^{r-1}} \end{align*}
  2. \begin{align*} && \frac1{1-x} &= \sum_{r=1}^{\infty} x^{r-1} \\ \underbrace{\Rightarrow}_{\int} && -\ln (1-x) &= \sum_{r=1}^{\infty} \frac1{r} x^r \\ \underbrace{\Rightarrow}_{x = \frac12} && \ln 2 &= \sum_{r=1}^{\infty} \frac1{r } \times \frac{1}{ 2^{r}} \\ \Rightarrow && 2 \ln 2 &= \sum_{r=1}^{\infty} \frac1{r } \times \frac{1}{ 2^{r-1}} \\ \end{align*}
  3. \begin{align*} && (1-x)^{-1/2} &= 1 + \frac{(-\tfrac12)}{1!} (-x) +\frac{(-\tfrac12)(-\tfrac32)}{2!}(-x)^2 + \cdots \\ &&&= \sum_{r=0}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots \cdot (2r-1)}{2^rr!} x^r \\ \underbrace{\Rightarrow}_{x = \frac23} && \sqrt{3} &= \sum_{r=0}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots \cdot (2r-1)}{r!} \frac1{3^r} \\ &&&= 1 + \frac{1}{1!} \frac23 + \frac13 \sum_{r=2}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots \cdot (2r-1)}{r!} \frac1{3^{r-1}} \\ \Rightarrow && 3\sqrt{3}-5 &= \sum_{r=2}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots \cdot (2r-1)}{r!} \frac1{3^{r-1}} \\ \end{align*}

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