If-n-Z-show-that-k-1-n-ln-2-1-1-k-lt-1-

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}) \dots + \ln^{2} (1 + \frac{1}{10})

+ \frac{1}{11^{2}} + \frac{1}{12^{2}} + \dots + \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}) \dots + \ln^{2} (1 + \frac{1}{10})

+ \frac{1}{10} - \frac{1}{11} + \frac{1}{11} - \frac{1}{12} + \dots + \frac{1}{n - 1} - \frac{1}{n}

= 0.886300 + \frac{1}{10} - \frac{1}{n} < 0.986300 < 1 (q . e . d)