Where did this step come from? Problem on series converge/divergence.

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1. The original problem is this, but I'm just trying to figure out this one particular step!

Ʃ n!/(nn)
n=1

2. I'm at this step:

lim ln(n/n+1)/(1/n) = lnp
n→∞

BUT THEN my teacher wrote out the next step, and I'm like, "What the heck is that?!"

This is the next step:

lim \frac{\frac{1}{n/(n+1)}*\frac{(n+1)*1-n*1}{(n+1)^{2}}}{-1/n^{2}}
n→∞

Aaaaaagh! What the heck is going on?!
How did he get to this step? O_O

I am stuck wondering how I am supposed to figure out the limit at: lim\frac{ln(n/(n+1)}{1/n}!

=_= HELP PLEASE!
Thank you!
 
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Lo.Lee.Ta. said:
1. The original problem is this, but I'm just trying to figure out this one particular step!

Ʃ n!/(nn)
n=1

2. I'm at this step:

lim ln(n/n+1)/(1/n) = lnp
n→∞

BUT THEN my teacher wrote out the next step, and I'm like, "What the heck is that?!"

This is the next step:

lim \frac{\frac{1}{n/(n+1)}*\frac{(n+1)*1-n*1}{(n+1)^{2}}}{-1/n^{2}}
n→∞

Aaaaaagh! What the heck is going on?!
How did he get to this step? O_O

I am stuck wondering how I am supposed to figure out the lim it at: lim\frac{ln(n/(n+1)}{1/n}!

=_= HELP PLEASE!
Thank you!

He was applying the L'Hospital's Rule to a limit of indeterminate form 0/∞.
 
drawar said:
He was applying the L'Hospital's Rule to a limit of indeterminate form 0/∞.
##\displaystyle \lim_{n\to\,\infty} \left( \frac{\displaystyle \ln\left(\frac{n}{n+1}\right)}{\displaystyle \frac{1}{n}}
\right)\ \ ## is of the indeterminate form, 0/0 .
 
SammyS said:
##\displaystyle \lim_{n\to\,\infty} \left( \frac{\displaystyle \ln\left(\frac{n}{n+1}\right)}{\displaystyle \frac{1}{n}}
\right)\ \ ## is of the indeterminate form, 0/0 .

Oops, sorry, just ignore what I've said.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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