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I need to check for convergence here. Why is every example completely different and nothing which I've learned from the examples before helps me in the next? Hate it!

$$ \sum_{n\geq1} \frac{n^{n-1}}{n!} $$

so I tried the ratio test and I got

$$ \frac{1}{1+\frac{1}{n}} $$ which converges to $1$ which isn't $>$ or $<$ then $1,$ so that doesn't work.

which test will work and how can I transform the series?

Stefan Hamcke
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loop
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4 Answers4

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Hint: Note that you can write $$ \frac{n^{n-1}}{n!}=\frac{n\cdot n\cdots n}{2\cdot3\cdots n} $$ where the numerator contains $n-1$ copies. This can be written as $$ \frac{n^{n-1}}{n!}=\frac{n}{2}\cdot\frac{n}{3}\cdots\frac{n}{n}. $$ Can this possibly tend to $0$ as $n\rightarrow\infty$?

Nick Peterson
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  • It might be a little confusing with all the $n$'s floating around. I would note that your expression above is at least $n/2$. – robjohn Sep 10 '13 at 15:34
  • @robjohn I've elucidated it a bit; my intention was for him to figure that out himself, though! – Nick Peterson Sep 10 '13 at 15:38
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Ratio test:

$$\frac{(n+1)^n}{(n+1)!}\frac{n!}{n^{n-1}}=\left(1+\frac1n\right)^n\frac n{n+1}\xrightarrow[n\to\infty]{}e>1$$

so the series...

Added on request:

Since

$$\frac{k!}{(k-1)!}=k\;,\;\;\frac1{n^{n-1}}=\frac n{n^n}\;,\;\;\text{we get}$$

$$\frac{(n+1)^n}{(n+1)!}\frac{n!}{n^{n-1}}=\frac{(n+1)^n}{n^n}\frac{n\cdot n!}{(n+1)!}=\left(1+\frac1n\right)^n\frac n{n+1}\ldots$$

DonAntonio
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  • Can u "babystep" this for me please? to much going on there – loop Sep 10 '13 at 15:33
  • to fast. its so frustrating. totally confused. i somehow know how to transform n^(n-1) so something ^(n) but in this case there are too many hidden steps for somebody like me – loop Sep 10 '13 at 15:45
  • @DominiqueLüber, people usually expect students of this stuff, which is more or less first undergraduate year in university, can handle basic algebra: fractions, factorials, exponents, etc. ... and if some student has problems with this then he must work through it since this is high school stuff. – DonAntonio Sep 10 '13 at 15:47
  • thats what everbody's telling me. but I somehow studied for this exam almost 1 year now, and for me, almost 30 now, I think i'll never get good at this stuff since my level in school was very bad. but i passed almost all exams in College, so it should be fine somehow I guess .. – loop Sep 10 '13 at 15:51
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Why is every example completely different and nothing which I've learned from the examples before helps me in the next?

Because in mathematics you need to think in principles, not in rote repetition. Every problem is new, and you have to determine which of all the various facts/processes/algorithms/theorems you have learned to date can be applied to this specific problem. That is the nature of mathematics. It makes mathematics simultaneously hard, but exciting as h***.

John
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    I always suspected that mathematicians were secretly adrenaline junkies... – robjohn Sep 10 '13 at 15:30
  • yeah but my brain won't allow me this :) my brain wants to access different kind of "ways" I learned. I know what u trying to say here but I guess I just hope to pass this exam and let others do the math. – loop Sep 10 '13 at 15:34
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Let's look at the more general function $\sum_{n\geq1} x^n\frac{n^{n-1}}{n!}$

The ratio of consecutive terms is

$\begin{align} x\frac{\frac{(n+1)^n}{(n+1)!}}{\frac{n^{n-1}}{n!}} &=x\frac{(n+1)^n}{(n+1)!}\frac{n!}{n^{n-1}}\\ &=x\frac{(n+1)^n}{(n+1)n^{n-1}}\\ &=x\frac{(n+1)^{n-1}}{n^{n-1}}\\ &= x(1+\frac1{n})^{n-1}\\ &\to xe\\ \end{align} $

For this to converge, it is necessary that $|xe| \le 1$, or $|x| \le \frac1{e}$.

Since $x = 1$, the sum diverges.

marty cohen
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  • also. to fast steps. how do you guys all know to transform this so fast? from step 2 to 3 or 3 to 4. dont know how you transform it on step3: you take one (n+1) out? so you have (n+1) * (n+1)^n-1? and then you can shorten the term? – loop Sep 10 '13 at 15:40
  • If he had written the denominators on the RHS of the first line in the other order, it would have been clearer. There's really no choice how to proceed: just group similar terms and cancel what is the same in both. – Ryan Reich Sep 10 '13 at 15:45
  • ok. Step 3 to 4 might be

    $$ \frac{n * (1+\frac{1}{n})^{n-1}}{n*1^{n-1}} $$

    or am I wrong?

     – loop Sep 10 '13 at 15:57