If n∈Z^+ , show that Σ_(k=1) ^n ln^2 (1+(1/k))<1


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Question Number 107481 by ZiYangLee last updated on 11/Aug/20

$If n \in Z^{+}, show that \underset{k = 1}{\sum^{n}} \ln^{2} (1 + \frac{1}{k}) < 1$
	Answered by 1549442205PVT last updated on 11/Aug/20

	$\underset{k = 1}{\sum^{n}} \ln^{2} (1 + \frac{1}{k}) < 1$ $We have \ln (1 + x) < x \forall x > 0 . Indeed,$ $Consider the function f (x) = x - \ln (1 + x)$ $f^{'} (x) = 1 - \frac{1}{1 + x} = \frac{x}{1 + x} > 0 \forall x > 0$ $\Rightarrow f (x) is increase on (0; + \infty)$ $Hence f (x) ⩾ f (0) = 0 \forall x \in [0; + \infty)$ $\Leftrightarrow x > \ln (1 + x) \forall x (0; + \infty) (q . e . d)$ $Hence, \ln (1 + \frac{1}{k}) < \frac{1}{k} \Rightarrow \ln^{2} (1 + \frac{1}{k}) < \frac{1}{k^{2}}$ $\frac{1}{{(k + 1)}^{2}} < \frac{1}{k (k + 1)} = \frac{1}{k} - \frac{1}{k + 1} . Therefore$ $\underset{k = 1}{\sum^{n}} \ln^{2} (1 + \frac{1}{k}) = \ln^{2} (2) + \ln^{2} (\frac{3}{2}) + \ln^{2} (\frac{4}{3})$ $+ \ln^{2} (\frac{5}{4}) + \ln^{2} (\frac{6}{5}) . . . + \ln^{2} (1 + \frac{1}{10})$ $+ \frac{1}{11^{2}} + \frac{1}{12^{2}} + . . . + \frac{1}{n^{2}}$ $< \ln^{2} (2) + \ln^{2} (\frac{3}{2}) + \ln^{2} (\frac{4}{3}) + \ln^{2} (\frac{5}{4}) + \ln^{2} (\frac{6}{5}) . . . + \ln^{2} (1 + \frac{1}{10})$ $+ \frac{1}{10} - \frac{1}{11} + \frac{1}{11} - \frac{1}{12} + . . . + \frac{1}{n - 1} - \frac{1}{n}$ $= 0.886300 + \frac{1}{10} - \frac{1}{n} < 0.986300 < 1 (q . e . d)$
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