Prove-that-i-1-n-1-i-2-i-gt-ln-n-1-

Question Number 184062 by CrispyXYZ last updated on 02/Jan/23

Prove that

\underset{i = 1}{\sum^{n}} \frac{1}{\sqrt{i^{2} + i}} > \ln (n + 1)

Answered by mr W last updated on 02/Jan/23

t h i n g s t o k n o w :

1)

\frac{a + b}{2} ⩾ \sqrt{a b}

\frac{1}{\sqrt{a b}} ⩾ \frac{2}{a + b}

2)

f (x) = 2 (\sqrt{x + 1} - 1) - \ln (x + 1)

f (0) = 0

f^{'} (x) = \frac{1}{\sqrt{x + 1}} (1 - \frac{1}{\sqrt{x + 1}}) > 0

i t m e a n s f o r x ⩾ 0 f (x) i s s t r i c t l y

i n c r e a s i n g, i . e . f (x) > 0

\Rightarrow 2 (\sqrt{x + 1} - 1) > \ln (x + 1)

\underset{i = 1}{\sum^{n}} \frac{1}{\sqrt{i^{2} + i}}

= \underset{i = 1}{\sum^{n}} \frac{1}{\sqrt{i (i + 1)}}

> \underset{k = 1}{\sum^{n}} (\frac{2}{\sqrt{i} + \sqrt{i + 1}}) s e e 1) a b o v e

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

= 2 (\sqrt{n + 1} - 1)

> \ln (n + 1) ✓ s e e 2) a b o v e

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