About Convergence Status Problem

Question

Can I get help for this question? I have no idea. $$\sum_{n=1}^\infty = \frac {(n^n)}{(n)!n^n}$$ The problem wants convergence status from me.

Why don't you cancel $n^n$ from numerator and denominator. You just have $\sum \frac{1}{n!}$ — Sahiba Arora, Jun 30 '17 at 14:27
Hint: $\frac{n^n}{n^n}=1$ and $\sum_{n\geq 1}\frac{1}{n!}=e-1$. — Jack D'Aurizio, Jun 30 '17 at 14:27
Ah didn't see that. I don't have any idea, started in this week. — Götebakar, Jun 30 '17 at 14:27

score 1 · Answer 1 · answered Jun 30 '17 at 14:28

1

Hint: simplifying the fraction gives $\sum_{n = 1}^{\infty}\frac{1}{n!} < 1 + 1/2 + 1/6 + \sum_{n = 4}^{\infty}\frac{1}{n^2}$.

answered Jun 30 '17 at 14:28

user388557

Ty for your hint but i still can't solve it. where am i missing – Götebakar Jun 30 '17 at 14:29
Do you know the integral test for convergence? – user388557 Jun 30 '17 at 14:29
Nope, dunno :(( – Götebakar Jun 30 '17 at 14:30
Then Jack D'Aurizio's hint is probably easier. – user388557 Jun 30 '17 at 14:31
I couldn't undertand that too idk why – Götebakar Jun 30 '17 at 14:32
What methods do you know about proving convergence? – user388557 Jun 30 '17 at 14:33
I will try to understand your answer. – Götebakar Jun 30 '17 at 14:35

score 0 · Accepted Answer · answered Jun 30 '17 at 14:35

0

Your series is $\sum \frac{1}{n!}$. Let $a_n = \frac{1}{n!}$. Then $$\lim|\frac{a_{n+1}}{a_n}|=\lim \frac{n!}{(n+1)!}=\lim\frac{1}{n+1}=0<1$$

By ratio test, the given series converges.

answered Jun 30 '17 at 14:35

Sahiba Arora

@Götebakar You can accept the answer if it satisfies you. – Sahiba Arora Jun 30 '17 at 14:37
Yh, sure. I will – Götebakar Jun 30 '17 at 14:37

2 Answers2