Consider the series $$\sum_{n=1}^{\infty} e^{-nx}$$ Why do we conclude that the series does not converge,if $x \leq 0$ ?
Because of the fact that $f_n(x)=e^{-nx} \to +\infty, \text{ if }x \leq 0 $ ?
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1It doesn't converge to $+\infty$ for $x = 0$. But that $e^{-nx}$ does not converge to $0$ for $x \leqslant 0$ is sufficient to conclude divergence. – Daniel Fischer May 11 '14 at 09:43
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1@DanielFischer do you mean it doesn't converge to $+\infty$ for $x\gt 0$? – Ellya May 11 '14 at 09:44
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1@ellya I meant $f_n(x) = e^{-nx}$ with "It", not the series. – Daniel Fischer May 11 '14 at 09:48
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ohh.. yes that is true.. – Ellya May 11 '14 at 09:52
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If a series $\sum u_n $converges, then $\lim_{n\to\infty} u_n = 0$. This is a basic result that you seem to be familiar with already.
Now if $x = 0$, then the limit in question is $1$. If $x < 0$, the limit is $\infty$. Therefore the series doesn't converge when $x \le 0$.
Ayman Hourieh
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Ok,I understand...And what happens,if $x>0$ ? Why is it then $0< e^{-nx}<1$? – evinda May 11 '14 at 09:45
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1@evinda because $e^{-nx}=1/e^{nx}$ and $e^{nx}$ is greater than 1 for $x\gt 0$ – Ellya May 11 '14 at 09:46
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1If $x > 0$ then $-x < 0$ and $e^{-x} < e^0 = 1$. The series in question becomes a convergent geometric series. – Ayman Hourieh May 11 '14 at 09:46
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Ok,I understand!!!Thank you both!!! And something else..so,if $\sum {u_n}$ converges $u_n \to 0$..but if $u_n \to 0$,can we conclude that $\sum u_n$ converges? – evinda May 11 '14 at 09:48
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@evinda No you can't make that conclusion. A standard counterexample is $\sum 1/n$. – Ayman Hourieh May 11 '14 at 09:51
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So,I can just say that if $u_n \not\to 0$,the series $\sum u_n$ does not converge.But,if $u_n \to 0$,I cannot tell if the series converges or not...right?? – evinda May 11 '14 at 09:53
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