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Could someone tell me the limit of this: $1 - (7/8)^3(11/12)^4(15/16)^5(19/20)^6$ etc. or $1 - \prod_{i=0}^\infty (\frac{4i+7}{4i+8})^{i+3}$ ? Thank you.

  • 1
    You mean product or sum? – 2017-02-15
  • 0
    I mean product. – 2017-02-15
  • 0
    A *number* does not converge — sequences of numbers do. – 2017-02-15

2 Answers 2

1

Starting as Ahmed S. Attaalla, but using a partial sum, we get this:

$\begin{array}\\ A_m &=\sum_{n=0}^{m} (n+3) \ln( \frac{4n+7}{4n+8})\\ &=\sum_{n=0}^{m} (n+3) \ln(1- \frac{1}{4n+8})\\ &=\sum_{n=0}^{m} (n+3) \ln(1- \frac{1}{4(n+2)})\\ &=\sum_{n=2}^{m} (n+1) \ln(1- \frac{1}{4n})\\ &<\sum_{n=2}^{m} -(n+1)(\frac1{4n}) \qquad\text{since } \ln(1-x) < -x\\ &=-\sum_{n=2}^{m} \frac{n+1}{4n}\\ & \to -\infty \text{ as } m \to \infty\\ \end{array} $

Therefore $P =\lim_{m \to \infty}e^{A_m} =0 $ so the limit is one.

0

Let,

$$P=(7/8)^3(11/12)^4....$$

Note then,

$$\ln P=3 \ln (7/8)+4 \ln (11/12)+....$$

Define $A=\ln P$ so that,

$$A=\sum_{n=0}^{\infty} (n+3) \ln( \frac{4n+7}{4n+8})$$

$$=\sum_{n=0}^{\infty} (n+3)(\ln (4n+7)-\ln (4n+8))$$

This diverges by the limit comparison test with $\sum_{n=1}^{\infty} (n+3)/n$, compare this with $\sum_{n=0}^{\infty} (n+3)(\ln (4n+8)-\ln (4n+7))$. All terms are negative, and we do not have an alternating sum, so this goes to $-\infty$ and thus your product must go to $0$. So the limit is $1$.