Let for a differentiable function f : ( 0 , ? ) R , f ( x ) - f ( y ) ? log e ( x y ) + x - y , ?x , y ? ( 0 , ? ) . Then ? n = 1 20 f ' ( 1 n 2 ) is equal to _____ . [2024]

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Solution

Q.
$Let for a differentiable function f : (0, \infty) \to R, f (x) - f (y) \geq \log_{e} (\frac{x}{y}) + x - y, \forall x, y \in (0, \infty) .$

$Then \sum_{n = 1}^{20} f' (\frac{1}{n^{2}}) is equal to _____ .$ [2024]

Ans.
(2890)

$f (x) - f (y) \geq \log_{e} (\frac{x}{y}) + x - y \forall (x, y) \in (0, \infty)$ ...(i)

Now, $f' (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \log_{e} \frac{(\frac{x + h}{x}) + h}{h}$

$= \lim_{h \to 0} [\frac{\log_{e} (1 + \frac{h}{x})}{h} + 1] = \frac{1}{x} + 1$ ..(ii)

Now, $\sum_{n = 1}^{20} f' (\frac{1}{n^{2}}) = \sum_{n = 1}^{20} \frac{1}{\frac{1}{n^{2}}} + 1$ [Using (i)]

$= \sum_{n = 1}^{20} (n^{2} + 1) = {[\frac{n (n + 1) (2 n + 1)}{6} + n]}_{n = 20}$

$= \frac{20 \times 21 \times 41}{6} + 20 = 2870 + 20 = 2890$

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