Math Problems Easier: limits and continuity

In Number theory tutorialoffered by TutorVista will help you to solve problem related to number sequences, calculating limits and continuity of the sequence.

Topic : Limits and convergence of given number sequence.

Decrease in the value of the numbers in the sequence is said to be sequence convergence.

Question : Write the first five terms of the sequence, determine whether the sequence converges, and if so find its limit.

$(1 - \frac{1}{2}), (\frac{1}{2} - \frac{1}{3}), (\frac{1}{3} - \frac{1}{4}), (\frac{1}{4} - \frac{1}{5}), . . . . . .$

Solution :

General term will be given as ,
$(\frac{1}{n}-\frac{1}{n+1})<br />\\ a_ = \frac{1}{1} - \frac{1}{1+1} = 1 - \frac{1}{2} = \frac{2-1}{2} = \frac{1}{2} = 0.5 \\ a_2 = \frac{1}{2} - \frac{1}{3} = \frac{3-2}{2*3} = \frac{1}{6} = 0.167 \\a_3= \frac{1}{3} - \frac{1}{4} = \frac{4-3}{3*4} = \frac{1}{12} = 0.083 \\ a_4 = \frac{1}{4} - \frac{1}{5} = \frac{5-4}{5*4} = \frac{1}{20} = 0.05 \\ a_5 = \frac{1}{5} - \frac{1}{6} = \frac{6-5}{6*5} = \frac{1}{30} = 0.03$

So 0.5, 0.167, 0.083, 0.05, 0.03 Thus we can observe that the values goes on decreasing.
Hence the number sequence converges.

Nows lets find it's limit by ratio test

$\mathop{\lim}\limits_{n \to \infty}(\frac{a_{n+1}}{a_n}) \\ = \mathop{\lim}\limits_{n \to \infty}(\frac{\frac{1}{n+1} -\frac{1}{n+1+1}}{\frac{1}{n} -\frac{1}{n+1}}) \\ = \mathop{\lim}\limits_{n \to \infty}(\frac{\frac{n+2-(n+1)}{(n+1)(n+2)}}{\frac{n+1-n}{n(n+1)}}) \\ \mathop{\lim}\limits_{n \to \infty} (\frac{n+2-n-1}{(n+1)(n+2)} \div \frac{n+1-n}{n(n+1)}) \\ \mathop{\lim}\limits_{n \to \infty}(\frac{1}{(n+1)(n+2)}X\frac{n(n+1)}{1}) \\ \mathop{\lim}\limits_{n \to \infty}(\frac{n}{n+2}) \\ = \frac{\infty}{\infty+2} \\ = \frac{\infty}{\infty} \\ = 1<br />$

For more help contact math help or calculus help.

Math Problems Easier

Sunday, May 24, 2009

Question on Limits and Convergence of Number Sequence

My Blog List

Categories

Recent Post