Let $x_n=\frac{1}{3}+(\frac{2}{5})^2+(\frac{3}{7})^3+....(\frac{n}{2n+1})^2$ Is $(x_n) $cauchy sequence?

Question

Let $$x_n=\frac{1}{3}+\left(\frac{2}{5}\right)^2+\left(\frac{3}{7}\right)^3+\cdots+\left(\frac{n}{2n+1}\right)^n$$ Is $(x_n)$ Cauchy sequence ?

My work

for $n>m:$

$$\begin{align}|x_n-x_m|&=\left|\frac{1}{3}+\left(\frac{2}{5}\right)^2+\cdots+\left(\frac{n}{2n+1}\right)^n-\left(\frac{1}{3}+\left(\frac{2}{5}\right)^2+\cdots+\left(\frac{m}{2m+1}\right)^m\right)\right|\\ &=\left|\left(\frac{m+1}{2(m+1)+1}\right)^{m+1}+\cdots+\left(\frac{n}{2n+!}\right)^n\right|\end{align}$$ how to proceed from here

Answer 1

For $n > m$ we have

\begin{align} |x_n - x_m| &= \left(\underbrace{\frac{m+1}{2(m+1)+1}}_{\le \frac12}\right)^{m+1}+\cdots +\left(\underbrace{\frac{n}{2n+1}}_{\le \frac12}\right)^n \\ &\le \frac{1}{2^{m+1}} + \cdots + \frac{1}{2^n} \\ &= \frac{1}{2^{m+1}}\left( 1 + \frac12 + \cdots + \frac1{2^{n-m-1}}\right)\\ &= \frac{1}{2^{m+1}} \frac{2^{n-m}-1}{2-1}\\ &= \frac{2^{n-m}-1}{2^{m+1}} \xrightarrow{m\to\infty} 0 \end{align}

Therefore your sequence $(x_n)_n$ is Cauchy.

Answer 2

Hint. Note that for $k\geq 1$ $$0<\left(\frac{k}{2k+1}\right)^k=(2+1/k)^{-k}<2^{-k}.$$ Therefore the for $n>m\geq 1$, $$|x_n-x_m|=\sum_{k=m+1}^{n} \left(\frac{k}{2k+1}\right)^k\leq \sum_{k=m+1}^{\infty} 2^{-k}=\frac{1}{2^{m}}.$$