# Convergence/divergence of series

I'm having trouble determine whether the following series converge or diverge. I'm supposed to use the ratio test for these.

$$\sum\limits_{n = 1}^\infty {\frac{{n!}}{{n^n }}}$$

$$\sum\limits_{n = 1}^\infty {\frac{{n^2 }}{{e^n }}}$$

For the first one I tried:

$$\left| {\frac{{a_{n + 1} }}{{a_n }}} \right| = \left| {\frac{{\left( {n + 1} \right)!}}{{n^{n + 1} }}\frac{{n^n }}{{n!}}} \right| = \frac{{n + 1}}{n} \to 1$$ as n tends to infinity so wouldn't the ratio test be inconclusive because the limit is 1? The answer tells me that the limit is supposed to be exp(-1) which is less than one so the series converges. The answer for the second one is the same. So I try to manipulate the summations so that they look like the one for the exponential.

$$e^x = \sum\limits_{n = 0}^\infty {\frac{{x^n }}{{n!}}}$$

$$= \sum\limits_{n = 0}^\infty {\left( {\frac{{n!}}{{x^n }}} \right)^{ - 1} }$$

$$e^n = \sum\limits_{n = 0}^\infty {\left( {\frac{{n!}}{{n^n }}} \right)^{ - 1} }$$

But how would I find the limit of (a_n+1)/(a_n) to be exp(-1)? Can someone help me out?

Hurkyl
Staff Emeritus
Gold Member
You try checking your work for silly mistakes? It's one of the first things I do when I think that the answer I'm getting is wrong.

LeonhardEuler
Gold Member
Take another look at your ratio test work. You also made a mistake in the last line of your post, but it won't matter because you won't need those equations. Your overworking the variable n.

Ok I see where I went wrong, thanks for the help.

$$\left| {\frac{{a_{n + 1} }}{{a_n }}} \right| = \left| {\frac{{\left( {n + 1} \right)!}}{{\left( {n + 1} \right)^{n + 1} }}\frac{{n^n }}{{n!}}} \right| = \frac{{\left( {n + 1} \right)n^n }}{{\left( {n + 1} \right)\left( {n + 1} \right)^n }} = \left( {\frac{n}{{n + 1}}} \right)^n$$

$$= \left( {1 + \frac{{ - 1}}{{n + 1}}} \right)^n = \left[ {\left( {1 + \frac{{ - 1}}{{n + 1}}} \right)^{n + 1} } \right]^{\frac{n}{{n + 1}}} \to e^{ - 1}$$ as n tends to infinity.

I checked my working numerous times but for some reason I just kept on missing the n^n error when I replaced n by (n+1). By the way, what was my mistake in the last line of my previous post? I think it would be helpful to me if I know what it is. After all, I might need to use it again some day.

Edit: Nevermind my last sentence. I know what you mean now LE. Edit 2: Does the product of limits apply when a limit to infinity is taken? For finite a and continuous functions f and g, lim(x->a)(fg) = (lim(x->a)f)(lim(x->a)g) if I recall correctly. Does this apply if a approaches infinity?

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lurflurf
Homework Helper
Benny said:
Does the product of limits apply when a limit to infinity is taken? For finite a and continuous functions f and g, lim(x->a)(fg) = (lim(x->a)f)(lim(x->a)g) if I recall correctly. Does this apply if a approaches infinity?
Yes, when all three limits exist. Limits don't care about the value of a only what the function does. You can use substitution to reduce to finite a.
$$\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow 0}g\left(\frac{1}{x}\right)$$
where there exist a real number qwt such that f(x)=g(x) for all x>qwt.
and g(1/0+)=g(1/0-) should either exist, and they either exist of do not together.

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Hurkyl
Staff Emeritus
Gold Member
$$\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow 0^+}f\left(\frac{1}{x}\right)$$

actually. I see that mistake made very often!

lurflurf
Homework Helper
Hurkyl said:
$$\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow 0^+}f\left(\frac{1}{x}\right)$$

actually. I see that mistake made very often!
$$\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow 0}f\left(\frac{1}{x}\right)$$

It is true when both limits exist, perhaps I should have said when the limits exist, or for for suitable f.