Test for convergence of the series $\displaystyle \sum_{n=1}^{\infty} \frac{\sqrt{n}}{x^n\sqrt{n^2+1}}$ when $\displaystyle x>0$
I am confused, since 'x' is in the denominator. Otherwise it could be treated as a general power series
Test for convergence of the series $\displaystyle \sum_{n=1}^{\infty} \frac{\sqrt{n}}{x^n\sqrt{n^2+1}}$ when $\displaystyle x>0$
I am confused, since 'x' is in the denominator. Otherwise it could be treated as a general power series
Use the quotient test, You'll obtain that for $\displaystyle x>1$ the series is convergent if $\displaystyle x>1$ and divergent if $\displaystyle 0<x<1$ . For $\displaystyle x=1$ is divergent (compare for example with $\displaystyle \sum_{n\geq 1} 1/n^{1/2}$ ).
Fernando Revilla
If you were to let z= 1/x, it would be a power series:
$\displaystyle \displaytype \sum_{n=1}^\infty \frac{\sqrt{n}}{\sqrt{n^2+1}}z^n$
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").
Of course, the ratio test can be used here...
$\displaystyle \displaystyle \lim_{n \to \infty}\frac{a_{n+1}}{a_n} = \lim_{n \to \infty}\frac{\frac{\sqrt{n+1}}{x^{n+1}\sqrt{(n+1)^ 2+1}}}{\frac{\sqrt{n}}{x^n\sqrt{n^2+1}}}$
$\displaystyle \displaystyle = \lim_{n \to \infty}\frac{x^n\sqrt{n^2+1}\sqrt{n+1}}{x^{n+1}\sq rt{n^2 + 2n + 2}\sqrt{n}}$
$\displaystyle \displaystyle = \frac{1}{x}\lim_{n \to \infty}\sqrt{\frac{(n^2+1)(n+1)}{n(n^2+2n+2)}}$
$\displaystyle \displaystyle = \frac{1}{x}\lim_{n \to \infty}\sqrt{\frac{n^3+n^2+n+1}{n^3 + 2n^2 +2n}}$
$\displaystyle \displaystyle = \frac{1}{x}\lim_{n \to \infty}\sqrt{\frac{1 + \frac{1}{n} + \frac{1}{n^2} + \frac{1}{n^3}}{1 + \frac{2}{n} + \frac{2}{n^2}}}$
$\displaystyle \displaystyle = \frac{1}{x}\cdot 1$
$\displaystyle \displaystyle = \frac{1}{x}$.
The series converges when this limit is $\displaystyle \displaystyle <1$. So we need to evaluate the values of $\displaystyle \displaystyle x$ for which this ratio is $\displaystyle \displaystyle <1$, and since $\displaystyle \displaystyle x > 0$...
$\displaystyle \displaystyle \frac{1}{x} < 1$
$\displaystyle \displaystyle 1 < x$.
So the series is convergent when $\displaystyle \displaystyle x > 1$.
Since the ratio test fails when the limit is $\displaystyle \displaystyle 1$, you need to substitute $\displaystyle \displaystyle x=1$ into the original series and test the convergence of the series for that value.