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Evaluate the Sum

$$S=\frac{1}{4}+\frac{1.3}{4.6}+\frac{1.3.5}{4.6.8}+\cdots \infty$$

My try: We have the $n$ th term as

$$T_n=\frac{1.3.5. \cdots (2n-1)}{4.6.8 \cdots (2n+2)}$$ $\implies$

$$T_n=\frac{1.3.5. \cdots (2n-1)}{2^n \times (n+1)!}$$

$$T_n=\frac{(2n)!}{4^n \times n! \times (n+1)!}$$

$$T_n=\frac{\binom{2n}{n-1}}{n \times 4^n}$$

Any clue here?

Ekaveera Gouribhatla
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  • It's not $\infty$ as $T_n$ converges to $0$. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Aug 09 '18 at 14:24
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    I would write it like this: $$T_n = \dfrac{\dbinom{2n}{n}}{(n+1)4^n}$$ because the central binomial is asymptotically equivalent to $$\dbinom{2n}{n} \sim \dfrac{4^n}{\sqrt{\pi n}}$$ which makes $$T_n \sim \dfrac{1}{(n+1)\sqrt{\pi n}}$$ – SlipEternal Aug 09 '18 at 14:28
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    It seems the question is not evaluate $T_n$, but evaluate the infinite series $\sum_n T_n$ – GEdgar Aug 09 '18 at 14:32
  • $$S_n = S_{n-1} + T_n$$

    Solving the recurrence relation gives:

    $$S_n = 1 - \dfrac{1}{2} \dfrac{\dbinom{2(n+1)}{n+1}}{4^n} + C$$

    Since $S_1 = \dfrac{1}{4}$, we have $C = 0$, which gives:

    $$S_n = 1 - \dfrac{\dbinom{2(n+1)}{n+1}}{2\cdot 4^n}$$

    Asymptotically, we have:

    $$S_n \sim 1 - \dfrac{2}{\sqrt{\pi(n+1)}}$$

    Taking the limit, we find:

    $$\lim_{n \to \infty} S_n = 1$$

     – SlipEternal Aug 09 '18 at 14:51
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    Noticing that $T_n$ is the nth Catalan number divided by $4^n$, we have $$T_n = \frac{2}{\pi}\int_{0}^{1} x^n\sqrt{\frac{1-x}{x}},dx$$ Summing to $N$, we get: $$\begin{align} \sum^N_1 T_n &= \frac{2}{\pi}\int_{0}^{1}\sqrt{\frac{1-x}{x}}\sum^N_1 x^n,dx \ &= \frac{2}{\pi}\int_{0}^{1}\sqrt{\frac{1-x}{x}} \frac{x(x^N-1)}{x-1} ,dx \ &= -\frac{2}{\pi}\int_{0}^{1}\sqrt{\frac{x}{1-x}} (x^N-1) ,dx \end{align} $$ Let $N \rightarrow \infty$ so that $$ \sum^{\infty}1 T_n = \frac{2}{\pi}\int{0}^{1}\sqrt{\frac{x}{1-x}} ,dx $$ and this integral is easily evaluated to be $1$. – Malkin Aug 09 '18 at 15:07

1 Answers1

3

Hint: See Catalan Number with $x=\dfrac14$.

Nosrati
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  • According to Wolframalpha, the answer is 1: http://www.wolframalpha.com/input/?i=Sum%5BCatalannumber%5Bn%5D%2F4%5En,%7Bn,1,infinity%7D%5D – SlipEternal Aug 09 '18 at 14:32
  • Maple agrees sum(binomial(2*n,n)/(n+1)/4^n,n=1..infinity); yields $1$. – GEdgar Aug 09 '18 at 14:36