Prove that $u_{n} = \frac{1}{n(10n+1)}$ is convergent

Question

Question

Prove that $u_{n} = \frac{1}{n(10n+1)}$ is convergent

Attempt

$u_{n} = \frac{1}{n(10n+1)}$

$u_{n} = \frac{1}{n(10n+1)} = \frac{A}{n}+\frac{b}{10n+1}=\frac{1}{n}-\frac{1}{10n+1}$

$S_{n} = u_{1}+u_{2}+u_{3}+...+u_{n-1}+u_{n} =\frac{1}{1}-\frac{1}{11}+\frac{1}{2}-\frac{1}{21}+\frac{1}{3}-\frac{1}{31}+\frac{1}{4}...+\frac{1}{10n+1}-\frac{1}{n}+ \frac{1}{n} -\frac{1}{10n+1} = 1 - \frac{1}{10n+1}$

S = $ \lim_{n \to \infty} S_{n} = \lim_{n \to \infty}(1-\frac{1}{10n+1})=1 $, thus it is Convergent. But I've messed something up with A,B and I can't figure out what exactly

Answer 1

I think you made a mistake in your partial fraction decomposition of $\frac{1}{n(10n+1)}$ since $$\frac{1}{n}-\frac{1}{10n+1}=\frac{9n+1}{n(10n+1)}\neq \frac{1}{n(10n+1)}$$ Hint: Check your value for $b$

Answer 2

Your question asks if $u_n$ is convergent, instead of $\sum_{n=1}^\infty u_n$. Certainly, $u_n \xrightarrow{n\to\infty} 0$. Regardless, for the sum, consider the following hints. Let $$v_n := \frac{u_n}{10} = \frac{1}{10n(10n+1)}$$ Clearly, $\sum v_n$ converges if and only if $\sum u_n$ converges. For the convergence of $\sum v_n$, the idea is to show that a "bigger" sum converges. Observe that $$\sum_{n=1}^\infty \frac{1}{n(n+1)}$$ converges, and $$\left\{\frac{1}{10n(10n+1)}: n\in \mathbb N\right\} \subsetneq \left\{\frac{1}{n(n+1)}: n\in \mathbb N\right\}$$ Can you fill in the details?

Answer 3

A simple elementary proof that bounds the expression by something that telescopes.

For $n \ge 2$

$\begin{array}\\ \dfrac{1}{n(10n+1)} &=\dfrac1{10}\dfrac{1}{n(n+\frac1{10})}\\ &\lt\dfrac1{10}\dfrac{1}{n(n-1)}\\ &=\dfrac1{10}\left(\dfrac{1}{n-1}-\dfrac{1}{n}\right)\\ \end{array}\\ $

so

$\begin{array}\\ \sum_{n=2}^m\dfrac{1}{n(10n+1)} &\lt \sum_{n=2}^m\dfrac1{10}\left(\dfrac{1}{n-1}-\dfrac{1}{n}\right)\\ &= \dfrac1{10}\left(1-\dfrac{1}{m}\right)\\ &< \dfrac1{10}\\ \end{array}\\ $