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
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 .. ?
Hello, alexandrabel90!
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: .$\displaystyle S_{n+1} \;>\;S_n \quad\hdots\quad S_n\text{ is an increasing function.}$
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