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.
On
If we take $\,x<0\,$ then everything's fine, since putting $\,y=-x>0\,$ and taking $\,n\,$ big enough:
$$e^{\sqrt n\,x}<n^{-2}\Longleftrightarrow n^2<e^{\sqrt n\,y}\Longleftrightarrow 2\log n<\sqrt n\,y$$
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.
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!