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I've to study this series:

$\sum_{n=1}^\infty e^{\sqrt n\,x}$

My teacher wrote that with the asymptotic comparison with this series:

$\sum_{n=1}^\infty\frac{1}{n^2}$
My series converges for every

$x<0$

I don't understand the motivation, hoping for someone to enlighten me!

=) Thanks. Leonardo.

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    Yes, yes, you are right, I'm going to correct! Thanks, but I don't understand again =D2012-11-18

2 Answers 2

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I think you may have copied the series down wrong, since $\displaystyle \sum_{n=1}^\infty e^{\sqrt{n}x}$ does not converge for $x>0$! Certainly $e^{\sqrt{n}x}$ is greater than one for any $n, x >0$ so the sum definitely diverges. I think you meant to say that it converges for $x < 0$ (or change a sign in the exponent in the sum), which is true.

To do this, we look at the power series expansion of $e^{\sqrt{n}x}$ and compare this term to $\frac{1}{n^2}$. For $x < 0$ we can write $x = - y$ where $y>0$, so $e^{\sqrt{n}x} = e^{-\sqrt{n}y}$. By looking at the fifth term in the series expansion of $e^{\sqrt{n}y}$, which is $\frac{n^2y^4}{4!}$, we can say $e^{\sqrt{n}y} >\frac{n^2y^4}{4!}$ and hence $e^{\sqrt{n}x} = e^{-\sqrt{n}y} <\frac{4!}{n^2y^4} = \frac{4!}{n^2x^4} $. So now we can compare our original series to a constant multiple ($\frac{4!}{x^4}$) of $\displaystyle\sum_{n=1}^\infty \frac{1}{n^2}$ to see that it converges!

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    @PizziraniLeonardo Great! Glad I could help.2012-11-18
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If we take $\,x<0\,$ then everything's fine, since putting $\,y=-x>0\,$ and taking $\,n\,$ big enough:

$e^{\sqrt n\,x}

Now you can check, for example with L'Hospital's Rule, that

$\lim_{w\to\infty}\frac{\log w}{\sqrt w\,y}=0$

so we're done.

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    You are right, thanks =)2012-11-19