Q4Mathematics I

Question

Find whether the following series is convergent or not?

\frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \frac{1}{4 \cdot 5} + \dots

Answer 1

To determine the absolute convergence of the given infinite series, we must first correctly identify the general $n$ -th term, which is mathematically given as:

u_{n} = \frac{1}{( n + 1 ) ( n + 2 )}

For massive values of $n$ , the constants $1$ and $2$ become mathematically negligible, meaning the term behaves asymptotically similar to $\frac{1}{n \cdot n} = \frac{1}{n ^{2}}$ . Therefore, to execute the Limit Comparison Test, we strategically choose the standard auxiliary comparison series $v_{n}$ to be:

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

We now rigorously evaluate the limit of the ratio $\frac{u _{n}}{v _{n}}$ as the index $n$ strictly approaches infinity:

n \to \infty lim \frac{u _{n}}{v _{n}} = n \to \infty lim \frac{\frac{1}{( n + 1 ) ( n + 2 )}}{\frac{1}{n ^{2}}} = n \to \infty lim \frac{n ^{2}}{( n + 1 ) ( n + 2 )}

Dividing both the numerator and the denominator by $n^{2}$ , we calculate the limit:

n \to \infty lim \frac{1}{( 1 + \frac{1}{n} ) ( 1 + \frac{2}{n} )} = \frac{1}{( 1 + 0 ) ( 1 + 0 )} = 1

Since this calculated limit is exactly $1$ (which is a strictly finite and non-zero positive number), the Limit Comparison Test firmly dictates that both series must behave identically—either both converge or both diverge.

We know from standard calculus theorems that the chosen auxiliary series $v_{n} = \frac{1}{n ^{p}}$ is a $p$ -series with $p = 2$ . Since $2 > 1$ , the $p$ -series test strictly dictates that $\sum v_{n}$ is convergent. Therefore, by the absolute mathematical rules of the Limit Comparison Test, the original given series $\sum u_{n}$ is undeniably convergent.

Q4Mathematics I

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