Sum of uneven unit fractions [closed]

Solution 1:

Ooh! I'm pretty proud of this proof! First step to this is let's simplify it down by a bit. Notice that the first 2 terms of the summation will always be: $\frac{1}{n+1}$ and $\frac{1}{n+3}$. Since all following terms are guaranteed to be positive, if we can prove that: $\frac{1}{n} < \frac{1}{n+1} + \frac{1}{n+3}$ for all even n, then we can prove you're original inequality is true for all even n. With this, let's try to prove the new simpler inequality by simplifying:

$\frac{1}{n} < \frac{1}{n+1} + \frac{1}{n+3}$

Multiply each fraction by the denominator of the other:

$\frac{1}{n} < \frac{n+3}{(n+3)(n+1)} + \frac{n+1}{(n+1)(n+3)}$

Combine fractions and expand the denominator:

$\frac{1}{n} < \frac{2n+4}{n^2+4n+3}$

Now multiply both sides of the inequality by n:

$1 < \frac{2n^2+4n}{n^2+4n+3}$

Now split that top $2n^2$ into $n^2 + n^2$:

$1 < \frac{n^2+4n+n^2}{n^2+4n+3}$

Now it's pretty clear. The first 2 terms in the numerator and denominator are the same, so as long as $n^2 > 3$, the numerator will be be larger than the denominator, and the inequality will be true. So, as a matter of fact, you are correct in your original statement, but the inequality is actually true for all $n \ge 2$.

Solution 2:

we first prove for even $n$ that: $$\frac{1}{n} < \frac{1}{n+1} + \frac{1}{n+3}$$ when $n=2$, $$\frac{1}{2}<\frac{1}{3}+\frac{1}{5}$$ when $n\geq 4$, $$\frac{1}{n} = \frac{1}{2n} + \frac{1}{2n} < \frac{1}{n+1} + \frac{1}{n+3}$$ so $$\frac{1}{n} < \frac{1}{n+1} + \frac{1}{n+3}\leq \sum_{i=0}^{n-1}\frac{1}{n+(2i+1)}$$