Prove that sequence $a_n/n$ is convergent for $a_n = 1+\frac15+\frac19... + \frac1{4n-3}$

Question

Show that the sequence $\langle a_n\rangle$ where

$a_n = 1+\frac{1}{5}+\frac{1}{9}... + \frac{1}{4n-3}$

does not converge, whereas $\langle b_n\rangle$ defined by $b_n = \frac{1}{n} a_n$ converges to 0.

My Approach: I was able to prove that $\langle a_n\rangle$ does not converge by showing that it is not a Cauchy sequence but not able to prove that $\langle b_n\rangle$ is converging sequence. Is there a theorem related to this?.

Answer 1

I completely misread the question at first. Let's try again. We know that for $k\geq1$, we have $4k-3\geq k$, and hence $$ \sum_{k=1}^n\frac{1}{4k-3}\leq \sum_{k=1}^n\frac{1}{k}. $$ The sum on the right gives the harmonic numbers, $H_n$, which are approximated asymptotically by $\ln(n)$ (given by a Riemann approximation of $\int_1^n\frac{1}{x}dx$). Thus since $$\lim_{n\to\infty}\frac{\ln(n)}{n}=0,$$ we may apply the squeeze theorem to $$0\leq \frac{1}{n}\sum_{k=1}^n\frac{1}{4k-3}\leq \frac{C+\ln(n)}{n}$$ (where $C$ is an arbitrarily large, fixed constant to make the inequality hold). Thus we have $b_n\to0$.

Answer 2

With Stolz–Cesàro theorem, $b_{n}=n$ is strictly monotone and divergent sequence and $${\displaystyle \lim _{n\to \infty }{\frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}} } ={\displaystyle \lim _{n\to \infty }\dfrac{1}{4n+1}}=0$$ Then, the limit $\lim_{n\to\infty}\dfrac{a_n}{n}=0$ as well.

Answer 3

Let, $$a_n=\sum_{k=1}^{n}\alpha_k$$ $$b_n=\sum_{k=1}^{n}\theta_k$$ In the first sequence, $$\alpha_k=\frac{1}{4k-3}$$ $$\theta_k=\frac{1}{n}\frac{1}{4k-3}$$ $$\theta_k=\frac{\alpha_k}{n}$$ Now taking derivative, $$\frac{d}{dk}[\theta_k]=\frac{-4}{n(4k-3)^2}$$ The derivative of $\theta_k$ is therefore negative for all $k$ such that $k\not= \frac{3}{4}$ is negative and for all $n$

Hence the $\theta_k$ is monotone decreasing in $k\gt 1$

Hence we can apply the integral test on $b_n$, $$\lim_{n\to \infty}\frac{1}{n}\int_1^{n}\frac{1}{4x-3} dx $$ $$\lim_{n\to \infty}\frac{\ln (4n-3)}{4n}-0$$ Applying L'Hospital rule, $$\lim_{n\to \infty}\frac{\frac{4}{4n-3}}{4}$$ $$\lim_{n\to \infty}\frac{1}{4n-3}$$ $$0$$

The integral is finite, hence the sum is convergent.

Answer 4

$$1+1/5+1/9+...\geq 1/2+1/6+1/10+...=$$ $$=(1/2)(1+1/3+1/5+...)\geq (1/2)(1/2+1/4+1/6+...)=$$ $$=(1/4)(1+1/2+1/3+...).$$ Let $s_n=\sum_{j=1}^n1/j.$ We have $$s_2=3/2.$$ $$s_4=s_2+(1/3+1/4)>s_2+2(1/4)=2.$$ $$s_8=s_4+(1/5+...+1/8)>s_4+4(1/8)>2+1/2=5/2.$$ $$s_{16}=s_8+(1/9+...+1/16)>s_8+8(1/16)>5/2+1/2=3.$$ Et cetera.