A-1-2018-1-2019-1-2050-find-the-integer-part-of-1-A-

Question Number 178777 by mr W last updated on 21/Oct/22

A = \frac{1}{2018} + \frac{1}{2019} + \dots + \frac{1}{2050}

f i n d t h e i n t e g e r p a r t o f \frac{1}{A} .

Commented by Frix last updated on 21/Oct/22

\frac{1}{A} should be close to \frac{2018 + 2050}{2 \times 33} = 61.6363 \dots

\Rightarrow answer should be 61

Commented by mr W last updated on 21/Oct/22

h o w i s \frac{1}{A} c l o s e t o \frac{2018 + 2050}{2 \times 33} ?

w e n e e d s t r i c t p r o o f .

Commented by Mathematification last updated on 21/Oct/22

Workings please

Commented by Frix last updated on 22/Oct/22

{it}^{'} s just an idea \dots

A = \frac{1}{n} + \frac{1}{n + 1} + \dots + \frac{1}{n + k - 1}

\Rightarrow

\frac{k}{A} is the harmonic mean of {n, n + 1, \dots, n + k - 1}

if q = \frac{n + k - 1}{n} is close to 1 \Leftrightarrow n is big enough and

k is not too big \frac{k}{A} is very close to \frac{n + (n + k - 1)}{2}

I cannot prove this . just tried these :

n = 10; k = 10; q = 1.9; \frac{k}{A} \approx 13.91; \frac{10 + 19}{2} = 14.5 g o o d

n = 100; k = 10; q = 1.09; \frac{k}{A} \approx 104.42; \frac{100 + 109}{2} = 104.5 b e t t e r

n = 10; k = 100; q = 10.9; \frac{k}{A} \approx 40.91; \frac{10 + 109}{2} = 59.5 b a d

so I guessed that with

n = 2018; k = 33; q = 1.016 {we}^{'} d get \frac{k}{A} close

to \frac{2018 + 2050}{2} = 2034 and \frac{1}{A} close to \frac{2034}{33}

which is \approx 61.64

Commented by mr W last updated on 22/Oct/22

t h a n k s f o r t h e e x p l a n a t i o n s i r!

Commented by Tawa11 last updated on 22/Oct/22

Great sir

Answered by mr W last updated on 22/Oct/22

A = \underset{18 n u m b e r s}{(\frac{1}{2018} + \frac{1}{2019} + \dots + \frac{1}{2035})} + \underset{15 n u m b e r s}{(\frac{1}{2036} + \dots + \frac{1}{2050})}

A > \frac{18}{2035} + \frac{15}{2050} = \frac{15 \times 2035 + 18 \times 2050}{2050 \times 2035}

\Rightarrow \frac{1}{A} < \frac{2050 \times 2035}{15 \times 2035 + 18 \times 2050} = \frac{166870}{2697}

< \frac{167214}{2697} = 62

A = \underset{33 n u m b e r s}{\frac{1}{2018} + \frac{1}{2019} + \dots + \frac{1}{2050}}

A < \frac{33}{2018}

\Rightarrow \frac{1}{A} > \frac{2018}{33} > \frac{2013}{33} = 61

s i n c e 61 < \frac{1}{A} < 62, t h e i n t e g e r p a r t o f

\frac{1}{A} i s 61 .

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