Show 1/(n+1)+1/(n+2)+1/(n+3)+1/2n>3/5, for n>2. Use mathematical induction.
THANKS
I think the statement is correct. To prove it by induction, write the nth and (n+1)st sums: $\displaystyle s_n=1/(n+1)+\dots+1/(2n)$; $\displaystyle s_{n+1}=1/(n+2)+\dots+1/(2n+2)$. Then express $\displaystyle s_{n+1}$ as $\displaystyle s_n+f(n)$ where $\displaystyle f(n)$ is some expression of n, i.e., $\displaystyle f(n)=s_{n+1}-s_n$. If you show that $\displaystyle f(n)>0$, you are done.
Note that $\displaystyle f(n)$ has three terms, and two of them are at least as big as half of the third.
Do you want to sum all terms from $\displaystyle \frac{1}{n+1}$ up to $\displaystyle \frac{1}{2n}$
not just the 4 terms quoted?
F(3)
$\displaystyle \frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{9}{20}+\ frac{1}{6}=\frac{37}{60}>\frac{36}{60}$
True for n=3
F(k)
$\displaystyle \frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}+.....+\f rac{1}{2k}>\frac{3}{5}$ ?
F(k+1)
$\displaystyle \frac{1}{k+2}+\frac{1}{k+3}+...+\frac{1}{2k}+\frac {1}{2k+1}+\frac{1}{2k+2}$
$\displaystyle =F(k)+\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{1}{k+1}$
$\displaystyle =F(k)+\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{2}{2k+2}$
$\displaystyle =F(k)+\frac{1}{2k+1}-\frac{1}{2k+2}$
A greater value in the denominator makes a smaller fraction, hence
$\displaystyle \frac{1}{2k+1}-\frac{1}{2k+2}>0$
Therefore if F(k) is true, F(k+1) certainly is