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Question Number 36218 by Rio Mike last updated on 30/May/18

$proof that$ $2 (\sqrt{n} - 1) < 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + . . . +$ $\frac{1}{\sqrt{n}} < 2 \sqrt{n}$
Commented byabdo mathsup 649 cc last updated on 30/May/18
$the function f(x)= (1/(√x)) is decreasing on ]0,+∞[so ∫_k ^(k+1) f(t)dt≤ f(k) ≤ ∫_(k−1) ^k f(t)dt ⇒ Σ_(k=1) ^n ∫_k ^(k+1) f(t)dt ≤ Σ_(k=1) ^n f(k) ≤ Σ_(k=1) ^n ∫_(n−1) ^n f(t)dt ⇒ ∫_1 ^(n+1) (dt/(√t)) ≤ 1 +(1/(√2)) +...+(1/(√n))≤ ∫_0 ^n (dt/(√t)) ⇒ [2(√t)]_1 ^(n+1) ≤ 1 +(1/(√2)) +....+(1/(√n))−≤ [2.(√n)]_0 ^n ⇒ 2{(√(n+1)) −1 }≤ 1 +(1/2) +...+(1/(√n)) ≤ 2(√n) but (√n_ ) ≤.(√(n+1)) ⇒ 2{(√n)−1} ≤ 1+(1/2) +....+(1/(√n)) ≤ 2(√n) .$
$t h e f u n c t i o n f (x) = \frac{1}{\sqrt{x}} i s d e c r e a s i n g o n] 0, + \infty [s o$ $\int_{k}^{k + 1} f (t) d t ⩽ f (k) ⩽ \int_{k - 1}^{k} f (t) d t \Rightarrow$ $\sum_{k = 1}^{n} \int_{k}^{k + 1} f (t) d t ⩽ \sum_{k = 1}^{n} f (k) ⩽ \sum_{k = 1}^{n} \int_{n - 1}^{n} f (t) d t \Rightarrow$ $\int_{1}^{n + 1} \frac{d t}{\sqrt{t}} ⩽ 1 + \frac{1}{\sqrt{2}} + . . . + \frac{1}{\sqrt{n}} ⩽ \int_{0}^{n} \frac{d t}{\sqrt{t}} \Rightarrow$ ${[2 \sqrt{t}]}_{1}^{n + 1} ⩽ 1 + \frac{1}{\sqrt{2}} + . . . . + \frac{1}{\sqrt{n}} - ⩽ {[2 . \sqrt{n}]}_{0}^{n} \Rightarrow$ $2 {\sqrt{n + 1} - 1} ⩽ 1 + \frac{1}{2} + . . . + \frac{1}{\sqrt{n}} ⩽ 2 \sqrt{n} b u t$ $\sqrt{n_{}} ⩽ . \sqrt{n + 1} \Rightarrow$ $2 {\sqrt{n} - 1} ⩽ 1 + \frac{1}{2} + . . . . + \frac{1}{\sqrt{n}} ⩽ 2 \sqrt{n} .$
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