P-1-1-3-1-5-1-7-1-2023-Q-1-1-2023-1-3-2021-1-5-2019-1-2023-1-P-Q-

Question Number 207661 by efronzo1 last updated on 22/May/24

P = 1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \dots + \frac{1}{2023}

Q = \frac{1}{1 \times 2023} + \frac{1}{3 \times 2021} + \frac{1}{5 \times 2019} + \dots + \frac{1}{2023 \times 1}

\frac{P}{Q} = ?

Answered by Frix last updated on 22/May/24

P (m) = \underset{k = 1}{\sum^{\frac{m + 1}{2}}} \frac{1}{2 k - 1} \land m = 2 n - 1

P (2 n - 1) = \underset{k = 1}{\sum^{n}} \frac{1}{2 k - 1} = H_{2 n - 1} - \frac{H_{n - 1}}{2}

Q (m) = \underset{k = 1}{\sum^{\frac{m + 1}{2}}} \frac{1}{(2 k - 1) (m + 1 - (2 k - 1)} \land m = 2 n - 1

Q (2 n - 1) = \underset{k = 1}{\sum^{n}} \frac{1}{(2 k - 1) (2 n - 2 k + 1)} =

= \frac{1}{2 n} (\underset{k = 1}{\sum^{n}} \frac{1}{2 k - 1} + \underset{k = 1}{\sum^{n}} \frac{1}{2 n - 2 k + 1}) =

[\underset{k = 1}{\sum^{n}} \frac{1}{2 k - 1} = \underset{k = 1}{\sum^{n}} \frac{1}{2 n - 2 k + 1}]

= \frac{1}{n} (H_{2 n - 1} - \frac{H_{n - 1}}{2})

\Rightarrow

\frac{P (2 n - 1)}{Q (2 n - 1)} = n

\Rightarrow

\frac{P (m)}{Q (m)} = \frac{m + 1}{2}

m = 2023 \Rightarrow Answer is 1012

Answered by MM42 last updated on 22/May/24

Q = \frac{1}{2024} [(1 + \frac{1}{2023}) + (\frac{1}{3} + \frac{1}{2021}) + (\frac{1}{5} + \frac{1}{2019}) + \dots + (\frac{1}{2019} + \frac{1}{5}) + (\frac{1}{2021} + \frac{1}{3}) + (\frac{1}{2023} + 1)]

= \frac{1}{2024} [2 p] = \frac{p}{1012}

\Rightarrow \frac{P}{Q} = 1012 ✓

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