Prove that $ \frac{1}{5}+ \frac{1}{9}+ \frac{1}{13}+\ldots+ \frac{ 1}{2005}< \frac{ 7}{4}$.

Question

Prove that $ \dfrac{1}{5}+ \dfrac{1}{9}+ \dfrac{1}{13}+\ldots+ \dfrac{ 1}{2005}< \dfrac{ 7}{4}$.

Solution $\sum_{k=1}^{501}\frac 1{4k+1}$ $=\frac 15+\sum_{k=2}^{501}\frac 1{4k+1}$ $<\frac 15+\int_1^{501}\frac{dx}{4x+1}$ $=\frac 15+\left[\frac{\ln(4x+1)}4\right]_1^{501}$

So $\sum_{k=1}^{501}\frac 1{4k+1}$ $<\frac 15+\frac{\ln 2005-\ln 5}4$ $=\frac 15+\frac{\ln 401}4$ $<\frac 74$ Q.E.D.

My questions:

Could you please post a simpler solution?

I just don't understand why $\frac 15+\frac{\ln 401}4<\frac 74$. Please explain why is that.

Thank you in advance!

Answer 1

This is not necessarily simpler because you have to know the upper bound on harmonic numbers $$H_n=\sum_{k=1}^n\frac{1}{k}\leq \ln{n}+\gamma+\frac{1}{2n}$$ where $\gamma\approx 0.577\ldots$, but then you can write $$\sum_{k=1}^{500}\frac{1}{4k+1}<\sum_{k=1}^{500}\frac{1}{4k}=\frac{1}{4}H_{500}\leq\frac{1}{4}(\ln{500}+\gamma+\frac{1}{1000})\approx 1.70\ldots<1.75$$

Answer 2

It is enough to invoke the Hermite-Hadamard inequality. Since $f(x)=\frac{1}{4x+1}$ is convex over $\mathbb{R}^+$,

$$ \int_{1/2}^{501+1/2}f(x)\,dx \geq f(1)+f(2)+\ldots+f(501) $$ hence $$ \sum_{k=1}^{501}\frac{1}{4k+1}\leq \frac{\log(669)}{4} $$ and it is enough to show that $e^7>669.$ On the other hand $e>2+\frac{2}{3}$ and $$\left(2+\frac{2}{3}\right)^7=2^7\cdot\frac{4}{3}\cdot\left(2-\frac{2}{9}\right)^3=\frac{2^{12}}{3}\left(1-\frac{1}{9}\right)^3>\frac{2^{13}}{3^2}>\frac{8000}{9}>888.$$