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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:

My series converges for every


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

=) Thanks. Leonardo.

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Something seems to be missing in what you wrote: $$e^{\sqrt n\,x}\rlap{\;\;\;\;/}\xrightarrow [n\to\infty]{} 0 $$ for $\,x>0\,$. It also would be a good idea if you checked in FAQ where they explain how to use LaTeX to properly write mathematics in this site. – DonAntonio Nov 18 '12 at 15:15
I know that doesn't go to zero, but that's exactly what my teacher wrote. (Ps thanks for the revision) – Pizzirani Leonardo Nov 18 '12 at 15:23
Well, then he's wrong. Perhaps he actually wrote $\,x<0\,$? – DonAntonio Nov 18 '12 at 15:25
Yes, yes, you are right, I'm going to correct! Thanks, but I don't understand again =D – Pizzirani Leonardo Nov 18 '12 at 15:29
up vote 1 down vote accepted

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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I've copied wrong! Thanks, I like your demonstration, but I have a question, for power series expansion you mean Taylor series expansion? Because we never did the one you mentioned =) Thanks again – Pizzirani Leonardo Nov 18 '12 at 15:38
Yeah, the power series expansion of $e^x$ is the taylor expansion. We actually define the function $e^x$ by syaing it is equal to $\displaystyle \sum_{i=0}^\infty \frac{x^i}{i!}$ and this then gives us that taylor expansion as being the same thing. The fifth term of this, which I mentioned in the post is $\frac{x^4}{4!}$ – Tom Oldfield Nov 18 '12 at 15:41
I've got it, I expand until fourth, so the square root can fade out and I can compare with $n^2$! Thanks a lot! – Pizzirani Leonardo Nov 18 '12 at 15:43
@PizziraniLeonardo Great! Glad I could help. – Tom Oldfield Nov 18 '12 at 15:45

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.

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My doubt is basically, why $e^{\sqrt{n}x}<n^{-2}$ ? – Pizzirani Leonardo Nov 18 '12 at 15:39
I just wrote it my answer! What's unclear? – DonAntonio Nov 18 '12 at 17:14
You started writing $e^{\sqrt{n}x}<n^{-2}$, but why? – Pizzirani Leonardo Nov 18 '12 at 18:20
This is logic: $$A\Longleftrightarrow B\Longleftrightarrow C$$ means "As iff B iff C", and this is what my first line means... – DonAntonio Nov 18 '12 at 19:03
You are right, thanks =) – Pizzirani Leonardo Nov 19 '12 at 19:13

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