Demonstrate that no matter n is , then : 1/(2n^2 + 1) + 1/(2n^2 + 2) +...+1/5n^2 > 3/4



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giorgiana1976's profile pic

Posted on (Answer #1)


=1/(2n^2+1)+...+1/3n^2 + 1/(3n^2+1)+...+1/4n^2 +1/(4n^2+1)+...+1/5n^2>

>1/3n^2 + ...+1/3n^2 + 1/4n^2+...+1/4n^2 + 1/5n^2+...



neela's profile pic

Posted on (Answer #2)

If  n is very large , we can write this as

Sn = summation (limit){(1/(1/2+1/n))(1/n)+(1/(2+2/n)0(1/n)+......(1/(5+0/n))(1/n) =

=Integral of (1/x)dx = {logx},x=2 tox=5 = log(5/2)=0.916291 >3/4.

otherwise if n is finite,  the the sum of the n terms is,

Sn =1/(2n^2+1) +1/(2n^2+2)+.......   ..+1/5n^2

On the right side there are 3n^2 terms. In the first n^2 terms 1/(3n^2+1) is least and  1/(4n^2+1) is the lowest valued term  in the 2nd n^2 terms and 1/5n^2 is the least  the 3rd n^2 terms. Therefore,

Sn>n^2 {(1/(3n^2) }+n^2{1/(4n^2+1)}+n^2{(1/5n^2)}

>{n^/3n^2}+{n^2/4n^2+1/5n^2) =(1/3+1/4+1/5) =47/60 which is higher than 3/4.

Thus for all values The inequality holds.




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