convergence of series with x^n in denominsator

**Sambit** · February 21st 2011, 06:17 AM

Test for convergence of the series $\sum_{n=1}^{\infty} \frac{\sqrt{n}}{x^n\sqrt{n^2+1}}$ when $x>0$

I am confused, since 'x' is in the denominator. Otherwise it could be treated as a general power series

**FernandoRevilla** · February 21st 2011, 06:32 AM

Use the quotient test, You'll obtain that for $x>1$ the series is convergent if $x>1$ and divergent if $0<x<1$ . For $x=1$ is divergent (compare for example with $\sum_{n\geq 1} 1/n^{1/2}$ ).

Fernando Revilla

**HallsofIvy** · February 21st 2011, 06:32 AM

If you were to let z= 1/x, it would be a power series:
$\displaytype \sum_{n=1}^\infty \frac{\sqrt{n}}{\sqrt{n^2+1}}z^n$
determine its interval of convergence and use that to determine for what values of x the original series converges (it son't be an "interval").

**Sambit** · February 21st 2011, 06:51 AM

For z=1/x, the series diverges for (-1,1), so for x, the series will converge for x<-1(ignore this, since x>0) and x>1. Thanks to HallsofIvy. Also at x=1, the convergence can be tested using comparison test. Thanks to FernandoRevilla.

**Prove It** · February 21st 2011, 07:41 AM

Of course, the ratio test can be used here...

$\displaystyle \lim_{n \to \infty}\frac{a_{n+1}}{a_n} = \lim_{n \to \infty}\frac{\frac{\sqrt{n+1}}{x^{n+1}\sqrt{(n+1)^ 2+1}}}{\frac{\sqrt{n}}{x^n\sqrt{n^2+1}}}$

$\displaystyle = \lim_{n \to \infty}\frac{x^n\sqrt{n^2+1}\sqrt{n+1}}{x^{n+1}\sq rt{n^2 + 2n + 2}\sqrt{n}}$

$\displaystyle = \frac{1}{x}\lim_{n \to \infty}\sqrt{\frac{(n^2+1)(n+1)}{n(n^2+2n+2)}}$

$\displaystyle = \frac{1}{x}\lim_{n \to \infty}\sqrt{\frac{n^3+n^2+n+1}{n^3 + 2n^2 +2n}}$

$\displaystyle = \frac{1}{x}\lim_{n \to \infty}\sqrt{\frac{1 + \frac{1}{n} + \frac{1}{n^2} + \frac{1}{n^3}}{1 + \frac{2}{n} + \frac{2}{n^2}}}$

$\displaystyle = \frac{1}{x}\cdot 1$

$\displaystyle = \frac{1}{x}$ .

The series converges when this limit is $\displaystyle <1$ . So we need to evaluate the values of $\displaystyle x$ for which this ratio is $\displaystyle <1$ , and since $\displaystyle x > 0$ ...

$\displaystyle \frac{1}{x} < 1$

$\displaystyle 1 < x$ .

So the series is convergent when $\displaystyle x > 1$ .

Since the ratio test fails when the limit is $\displaystyle 1$ , you need to substitute $\displaystyle x=1$ into the original series and test the convergence of the series for that value.

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