$\displaystyle
\left(\frac{k}{k + 1}\right)^{k^2}
$
The book answer is confusing to me - I would think you could just take the limit of
$\displaystyle
\frac{k}{k + 1}
$
but I am missing something apparently
$\displaystyle
\left(\frac{k}{k + 1}\right)^{k^2}
$
The book answer is confusing to me - I would think you could just take the limit of
$\displaystyle
\frac{k}{k + 1}
$
but I am missing something apparently
If you apply the Cauchy root test then you need to find the following limit:
$\displaystyle \lim_{k \rightarrow + \infty} \left( \frac{k}{k+1} \right)^k = \lim_{k \rightarrow + \infty} \left( 1 - \frac{1}{k+1} \right)^k$.
It is not too hard to show that the value of this limit is equal to $\displaystyle e^{-1} = \frac{1}{e}$.
re: bolded - maybe for you lol.
thank you mister fantastic
I will wrestle with it and also try to figure out why you need
$\displaystyle \lim_{k \rightarrow + \infty} \left( \frac{k}{k+1} \right)^k = \lim_{k \rightarrow + \infty} \left( 1 - \frac{1}{k+1} \right)^k$.
This problem definitely touches on some of my weaker areas. Also thanks for the latex fix, now I learned something new there too
1. $\displaystyle \left[ \left( \frac{k}{k+1}\right)^{k^2}\right]^{1/k} = \left( \frac{k}{k+1}\right)^{\frac{k^2}{k}} = \left( \frac{k}{k+1}\right)^{k}$.
2. $\displaystyle \frac{k}{k+1} = \frac{(k + 1) - 1}{k+1} = \frac{k+1}{k+1} - \frac{1}{k+1} = 1 - \frac{1}{k+1}$.
3. Use the well known limit from calculus: $\displaystyle \lim_{n \rightarrow + \infty} \left(1 + \frac{a}{n} \right)^n = e^a$.