# Convergence/divergence of series

• Benny
In summary, the conversation discusses using the ratio test to determine whether two series converge or diverge. The first series is inconclusive, but by manipulating the summations, it is found that the limit of (a_n+1)/(a_n) is e^-1. The conversation also mentions the product of limits and its applicability when taking a limit to infinity. The mistake of substituting n+1 for n in the ratio test is discussed and the correct approach is shown.

#### Benny

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?

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.

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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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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$$\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow 0^+}f\left(\frac{1}{x}\right)$$

actually. I see that mistake made very often!

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.