Question

Prove that x base n defined by xn =1/1.2 +1/2.3 +1/3.4 +,,,,,,+1/n(n+1) is convergent

Answer 1

Consider the given series as:

$x_{n}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{n(n+1)}$

So, Let $u_{n}=\frac{1}{n(n+1)}$ and $v_{n}=\frac{1}{n^{2}}$

By using the comparison test, which states

Let $\sum u_{n}$ and $\sum v_{n}$ be two series of positive terms and let $\sum v_{n}$ be convergent.  Then $\sum u_{n}$  converges, if

(a) If $\sum u_{n} < \sum v_{n}$ for all natural numbers

(b) $\frac{\sum u_{n}}{\sum v_{n}}\leq k$ where k is > 0 and finite.

(c) $lt\frac{ u_{n}}{v_{n}}=$ finite limit

Therefore,

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

Let $v_{n}=\frac{1}{n^{2}}$

Consider $lt_{n\rightarrow \infty }\frac{\sum u_{n}}{\sum v_{n}}=lt_{n\rightarrow \infty } \frac{1}{n(n+1)} \times n^{2}$

= $\frac{n}{n+1}$

= $lt _{n\rightarrow \infty }\frac{1}{1+\frac{1}{n}}$

= 1 (Finite)

Therefore, by comparison test, $\sum u_{n}$ and $\sum v_{n}$ converge and diverge together.

Since, $\sum v_{n}$= $\sum \frac{1}{n^{2}}$ which is convergent by p test as p = 2(>1).

p test states that:

The infinite series $\sum \frac{1}{n^{p}}$

(i) Convergent when p > 1 and Divergent when p ≤1.

Therefore, $\sum u_{n}$ also converges.