Two Infinite Series problems

**fizzle45** · March 25th 2010, 08:44 PM

Hi I am supposed to determine if the following series converges or diverges. The catch is that we are only allowed to use the comparison test, ratio test, and root test.
First one is
$\sum[\sqrt{n+1}-\sqrt{n}]$

and 2nd one is
$\sum\frac{n!}{n^n}$

any hints or the full blown solution would be great.

**Soroban** · March 25th 2010, 09:09 PM

Hello, fizzle45!

The second one has a fascinating solution!

We should know the definition of $e\!:\;\;\;e \;=\;\lim_{n\to\infty}\left(1 + \tfrac{1}{n}\right)^n$

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

Ratio test: . $\frac{a_{n+1}}{a_n} \;\;=\;\;\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n} {n!} \;\;=\;\;\frac{(n+1)!}{n!}\cdot\frac{n^n}{(n+1)^{n +1}}$

. . . . . . . . . $=\;\;\frac{n+1}{1}\cdot\frac{n^n}{(n+1)(n+1)^n} \;\;=\;\;\frac{n^n}{(n+1)^n}\;\;=\;\;\left(\frac{n }{n+1}\right)^n$

Divide numerator and denominator by $n\!:\;\;\;\left(\frac{1}{1+\frac{1}{n}}\right)^n \;\;=\;\;\frac{1}{\left(1 + \frac{1}{n}\right)^n }$

$\text{Take the limit: }\;\lim_{n\to\infty}\frac{1}{(1 + \frac{1}{n})^n} \;\;=\;\;\frac{1}{\underbrace{\lim(1+\tfrac{1}{n}) ^n}_{\text{This is }e}} \;\;=\;\;\frac{1}{e}$

Since the ratio is $\frac{1}{e}$ , which is less than 1, the series converges.

**skeeter** · March 26th 2010, 05:05 AM

Originally Posted by fizzle45

Hi I am supposed to determine if the following series converges or diverges. The catch is that we are only allowed to use the comparison test, ratio test, and root test.
First one is
$\sum[\sqrt{n+1}-\sqrt{n}]$

do this first, then determine what test is appropriate ...

$<br /> \left(\sqrt{n+1} - \sqrt{n}\right) \cdot \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}<br />$

**fizzle45** · March 26th 2010, 10:10 AM

here's what i came up with
$\left(\sqrt{n+1} - \sqrt{n}\right) \cdot \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}$
using root test
$\left(\frac{1}{\sqrt{n+1}+\sqrt{n}}\right)^\frac{1 }{n}=\frac{(1)^\frac{1}{n}}{\left(\sqrt{n+1}+\sqrt {n}\right)^\frac{1}{n}}$ i guess here is where i get stuck

i have to go to class right to turn this in, but i should that it diverges by comparing it to $\frac{1}{\sqrt{n+1}+\sqrt{n+1}}$ which can easily shown that it diverges.

**Krizalid** · March 26th 2010, 12:36 PM

find a lower bound, it diverges, and it's faster than applying any test.

**Laurent** · March 26th 2010, 02:23 PM

Originally Posted by fizzle45

Hi I am supposed to determine if the following series converges or diverges. The catch is that we are only allowed to use the comparison test, ratio test, and root test.
First one is
$\sum[\sqrt{n+1}-\sqrt{n}]$

Please, don't apply any test!

You have $\sum_{n=1}^N (\sqrt{n+1}-\sqrt{n})=(\sqrt{2}-\sqrt{1})+\cdots+(\sqrt{N+1}-\sqrt{N})=\sqrt{N+1}-1$ (every term but the extreme ones cancel : "telescopic sum"). Since $\sqrt{N+1}-1\to +\infty$ when $N\to\infty$ , the series diverges (cf. the very definition of convergence). This is WAY more elementary than any kind of test : just using the definition of convergence.

(if by any chance the square brackets [ ] mean "integer part", then every term in the sum is zero so the series converges)

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