How do you test the series Sigma n^2/2^n from n is [0,oo) for convergence?

Mar 2, 2017

the series converges

Explanation:

We can apply d'Alembert's ratio test:

Suppose that;

$S = {\sum}_{r = 1}^{\infty} {a}_{n} \setminus \setminus$, and $\setminus \setminus L = {\lim}_{n \rightarrow \infty} | {a}_{n + 1} / {a}_{n} |$

Then

if L < 1 then the series converges absolutely;
if L > 1 then the series is divergent;
if L = 1 or the limit fails to exist the test is inconclusive.

So our series is;

$S = {\sum}_{n = 0}^{\infty} {n}^{2} / {2}^{n}$

So our test limit is:

$L = {\lim}_{n \rightarrow \infty} | \frac{{\left(n + 1\right)}^{2} / {2}^{n + 1}}{{n}^{2} / {2}^{n}} |$
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} | {\left(n + 1\right)}^{2} / {2}^{n + 1} \cdot {2}^{n} / \left\{{n}^{2}\right\} |$
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} | {\left(n + 1\right)}^{2} / \left(2 \cdot {2}^{n}\right) \cdot {2}^{n} / \left\{{n}^{2}\right\} |$
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} | {\left(n + 1\right)}^{2} / 2 \cdot \frac{1}{{n}^{2}} |$
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} | \frac{1}{2} \cdot {\left(\frac{n + 1}{n}\right)}^{2} |$
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} | \frac{1}{2} \cdot {\left(1 + \frac{1}{n}\right)}^{2} |$
$\setminus \setminus \setminus = | \frac{1}{2} \cdot {\left(1 + 0\right)}^{2} |$
$\setminus \setminus \setminus = \frac{1}{2}$

and we can conclude that the series converges

Mar 2, 2017

This series converges

Explanation:

The easiest shot is the ratio test , so we look at:

abs((a_(n+1))/(a_n))_(n to oo

Here that means:

${\left\mid \frac{{\left(n + 1\right)}^{2} / \left({2}^{n + 1}\right)}{{n}^{2} / \left({2}^{n}\right)} \right\mid}_{n \to \infty}$

$= {\left\mid \frac{{\left(n + 1\right)}^{2}}{2 {n}^{2}} \right\mid}_{n \to \infty}$

$= {\left\mid \frac{1 + \frac{2}{n} + \frac{1}{n} ^ 2}{2} \right\mid}_{n \to \infty} = \frac{1}{2}$

So ${\left\mid \frac{{a}_{n + 1}}{{a}_{n}} \right\mid}_{n \to \infty} < 1$ which means that it converges absolutely.