may i know how do i show that Sn = n^2 /(n^2 +1) is increasing?

i tried ratio test and it gave an answer of 1..and i cant use comparision test since the denominator is not (n^2 -1).

thanks

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- May 2nd 2010, 08:27 AMalexandrabel90increasing sequence
may i know how do i show that Sn = n^2 /(n^2 +1) is increasing?

i tried ratio test and it gave an answer of 1..and i cant use comparision test since the denominator is not (n^2 -1).

thanks - May 2nd 2010, 08:32 AMMiss
How in earth can you show that the sequence is increasing by the comparison tests or the ratio test ??

Let $\displaystyle f(n)=S_n$

$\displaystyle \implies f(x)=\frac{x^2}{x^2+1}$

Differentiate it and try to remember how to show that the function is increasing by using its derivative ..

If you want to test its convergence (as an infinite series not a sequence), clearly $\displaystyle \lim_{n\to\infty} \frac{n^2}{n^2+1} \neq 0$ , so .. ? - May 2nd 2010, 09:14 AMalexandrabel90
i learnt in number theory that i can check if a sequence is increasing by whether the following value is greater than or smaller than the previous one.

sorry, i dont really get what you are trying to hint to me:/ - May 2nd 2010, 10:05 AMSoroban
Hello, alexandrabel90!

Quote:

How do i show that: .$\displaystyle S_n \:=\:\frac{n^2}{n^2 +1}$ .is increasing?

Compare $\displaystyle S_{n+1}$ to $\displaystyle S_n$

. . . . . . . . . $\displaystyle \frac{(n+1)^2}{(n+1)^2+1} \;\begin{array}{c}<\\ [-2mm] > \end{array} \;\frac{n^2}{n^2+1} $

. . . . . . .$\displaystyle (n^2+1)(n+1)^2 \;\begin{array}{c}< \\[-2mm] >\end{array}\;n^2[(n+1)^2+1] $

. . $\displaystyle n^4 + 2n^3 + 2n^2 + 2n + 1 \;\begin{array}{c}< \\[-2mm]>\end{array} \;n^4 + 2n^3 + 2n^2 $

. . . . . . . . . . . . .$\displaystyle 2n + 1 \;\begin{array}{c}< \\[-2mm]>\end{array}\;0 $

. . For $\displaystyle n \,\geq\,1,\;2n+1 \;\;{\color{red}>}\;\;0$

Therefore: .$\displaystyle S_{n+1} \;>\;S_n \quad\hdots\quad S_n\text{ is an increasing function.}$

- May 2nd 2010, 10:06 AMMiss
$\displaystyle f$ is increasing if and only if $\displaystyle f' > 0$ ..

- May 2nd 2010, 10:43 AMArchie Meade
Hi alexandrabel90,

if you are denoting the nth term of the sequence as Sn, then you can use the ratio test, without evaluating the limit.

Note that Sn normally denotes the sum of a series of terms of the sequence.

$\displaystyle \frac{S_{n+1}}{S_n}=\frac{\left(\frac{(n+1)^2}{(n+ 1)^2+1}\right)}{\left(\frac{n^2}{n^2+1}\right)}$

$\displaystyle =\frac{(n+1)^2}{(n+1)^2+1}\ \frac{n^2+1}{n^2}$

$\displaystyle =\frac{\left(n^2+2n+1\right)\left(n^2+1\right)}{\l eft(n^2+2n+2\right)n^2}$

$\displaystyle =\frac{\left(n^4+2n^3+2n^2\right)+2n+1}{n^4+2n^3+2 n^2}$

$\displaystyle =1+\frac{2n+1}{n^4+2n^3+2n^2}$

This is >1