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Mathematics 23 Online
OpenStudy (anonymous):

Convergence, divergence

OpenStudy (anonymous):

Determine if each series is a convergent series or a divergent series. |dw:1414130559264:dw|

OpenStudy (anonymous):

Would you use a root test, ratio test or integral test for this?

OpenStudy (freckles):

I was trying ratio test

OpenStudy (freckles):

but my cat took my ink pen

OpenStudy (freckles):

i got it back

OpenStudy (anonymous):

Haha, Hi freckles

OpenStudy (anonymous):

under what circumstance would you use an integral test

OpenStudy (freckles):

i would never use integral test if I had a factorial or alternating series

OpenStudy (freckles):

it would have to be something I could at least integrate

OpenStudy (anonymous):

true XD makes sense

OpenStudy (freckles):

root test also doesn't sound right either

OpenStudy (freckles):

because of the factorial business

OpenStudy (freckles):

i have you tried ratio test

OpenStudy (freckles):

crap seriously she keeps taking my pen

OpenStudy (freckles):

\[\lim_{n \rightarrow \infty}| \frac{(n+1)!}{(n+1)^{n+1}} \frac{n^n}{n!}|\]

OpenStudy (anonymous):

I'm working on it right now

OpenStudy (anonymous):

how would you simplify the n^n/(n+1)^n+1 part

OpenStudy (freckles):

this is the way i'm approaching it so far \[\lim_{n \rightarrow \infty}| \frac{(n+1)!}{(n+1)^{n+1}} \frac{n^n}{n!}| \\ = \lim_{n \rightarrow \infty}|\frac{n+1}{(n+1)^{n}(n+1)} n^n| \\ =\lim_{n \rightarrow \infty}|(\frac{n}{n+1})^n|\] \[=\lim_{n \rightarrow \infty}|e^{\ln((\frac{n}{n+1})^n)}|=\lim_{n \rightarrow \infty}|e^{n \ln(\frac{n}{n+1})}| \\ = \lim_{n \rightarrow \infty} | e^{\frac{\ln(\frac{n}{n+1})}{\frac{1}{n}}}| \]

OpenStudy (anonymous):

for line 2 how did you move the ^n+1

OpenStudy (freckles):

\[x^{2}=x^{1+1}=x^1 x^1\\x^{n+1}=x^{n}x^{1}\]

OpenStudy (freckles):

law of exponents

OpenStudy (freckles):

\[x^rx^s=x^{r+s}\]

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