use-limit-comparison-test-to-determine-the-series-converge-or-diverge-k-1-ln-1-1-k-2-

Question Number 69212 by aliesam last updated on 21/Sep/19

u s e l i m i t c o m p a r i s o n t e s t t o d e t e r m i n e

t h e s e r i e s c o n v e r g e o r d i v e r g e

\underset{k = 1}{\sum^{\infty}} l n (1 + \frac{1}{k^{2}})

Answered by mind is power last updated on 21/Sep/19

\forall x \in {I R}_{+} \ln (1 + x) ⩽ x

\Rightarrow \forall k \in N^{*} 0 ⩽ \ln (1 + \frac{1}{k^{2}}) ⩽ \frac{1}{k^{2}}

\Rightarrow \underset{k = 1}{\sum^{+ \infty}} \ln (1 + \frac{1}{k^{2}}) ⩽ \sum_{k = 1}^{+ \infty} \frac{1}{k^{2}} = \frac{π^{2}}{6}