Prove that $$\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{(n+1)^2} < n \cdot \left(1-\frac{1}{\sqrt[n]{2}}\right).$$
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Tags: inequalities, inequalities proposed, Kazakhstan
vi1lat
10.01.2015 16:24
Also we can use math. induction. method
leonardg
10.01.2015 16:28
vi1lat wrote: Also we can use math. induction. method Please , we all are curious...
sqing
23.01.2015 03:32
q32243 wrote: Prove that $\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{(n+1)^2} < n(1-\frac{1}{\sqrt[n]{2}})$ $\frac{1}{(n+1)^2} <\frac{1}{(n+\frac{1}{4})(n+\frac{5}{4})} .$ Strengthening Prove that\[\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{(n+1)^2} < n(1-\frac{1}{\sqrt[3n]{e^2}}).\]
Newbiee
16.04.2015 13:52
it not the national olimpiad of Kazakhstan, is a regional stage
Satvikgupta
08.12.2015 19:04
The given sum is less than the area under graph of yx^2=1 from x=1 to n. Using integration the problem is straightforward.